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Use these standard ultrafilter facts: a space is compact if and only if every ultrafilter on it converges; the image ultrafilter under a map is an ultrafilter; and a convergent ultrafilter contains every neighborhood of its limit.

Let $X={\prod }_{\lambda \in \Lambda }{X}_{\lambda }$ with all factors compact. An empty factor makes $X$ empty and hence compact. Otherwise, given an ultrafilter $\mathcal{U}$ on $X$, push it forward by each product projection to obtain an ultrafilter ${\mathcal{U}}_{\lambda }$ on $X_{\lambda }$. Choose a limit $x_{\lambda }$ of each, and put $x=(x_{\lambda }{)}_{\lambda }$. A basic product neighborhood of $x$ is an intersection of finitely many inverse images of coordinate neighborhoods. Each inverse image belongs to $\mathcal{U}$, so their finite intersection does too. Thus $\mathcal{U}$ converges to $x$, proving compactness of the product. Uniqueness of coordinate limits is not needed, so the factors need not be Hausdorff.

For local compactness, suppose only the factors indexed by a finite set $F$ may be noncompact. At $x\in X$ choose, for each $\lambda \in F$, a compact neighborhood $K_{\lambda }$ of $x_{\lambda }$, with an open neighborhood $U_{\lambda }\subseteq K_{\lambda }$. For other indices put $K_{\lambda }=X_{\lambda }$. The compact product ${\prod }_{\lambda }{K}_{\lambda }$ contains the open product neighborhood restricting just the finitely many coordinates in $F$ to $U_{\lambda }$. Hence every point has a compact neighborhood and the product is locally compact.

The unrestricted assertion is false. Each factor of ${\mathbb{R}}^{\mathbb{N}}$ is locally compact, but its product is not. Any basic neighborhood restricts only finitely many coordinates. Choose a free coordinate $j$; its projection of that neighborhood is all of $\mathbb{R}$. A compact neighborhood containing it would consequently have compact image equal to $\mathbb{R}$, impossible. This shows directly that no point has a compact neighborhood.

Write $\alpha X=X\cup \{{\infty }_X\}$. Its open sets are the open subsets of $X$ together with sets $\alpha X\setminus K$, where $K$ is a compact subset of $X$. Compact subsets of the Hausdorff space $X$ are closed. This fact, and finite unions of compact sets, verify the topology axioms, including intersections of an old open set with a neighborhood of infinity. The inclusion $h_1:X\to \alpha X$ is a topological embedding with open image.

An open cover of $\alpha X$ has a member containing infinity; its complement is compact in $X$ and therefore is covered by finitely many other members. This proves compactness. Two points of $X$ can be separated in $X$. A point $x$ and infinity can be separated by an open neighborhood $U$ of $x$ contained in a compact neighborhood $K$, and $\alpha X\setminus K$. Thus $\alpha X$ is Hausdorff. Since $X$ is noncompact, every neighborhood of infinity meets $X$, so the embedded copy of $X$ is dense. This proves it is a Hausdorff compactification.

For a proper map, use the standard locally compact Hausdorff convention: it is continuous and inverse images of compact sets are compact. DefineAn open subset of $Y$ has open inverse image in $X$, hence in $\alpha X$. A basic neighborhood $\alpha Y\setminus K$ of infinity has inverse image $\alpha X\setminus f^{-1}(K)$, open because $f^{-1}(K)$ is compact. Hence $\alpha f$ is continuous and the diagram commutes. Any continuous map from a compact space to a Hausdorff space is proper: a compact subset of the target is closed, its inverse image is closed in the compact domain and therefore compact. This proves the required properness.

The last requested extension does not exist, even as a continuous map. Both sequences $h_1(0,n)$ and $h_1(1,n)$ tend to ${\infty }_{{\mathbb{R}}^2}$, since they eventually leave every compact subset of the plane. Their prescribed images would be the constant sequences $h_2(0)$ and $h_2(1)$. Continuity would force both distinct points to equal $g({\infty }_{{\mathbb{R}}^2})$, contradicting uniqueness of limits in the Hausdorff compactification of the line. The original product projection is not proper, as its inverse image of a singleton is an unbounded line.

A T0 space distinguishes any two distinct points by an open set containing one and not the other. A Tychonoff space, also called a completely regular Hausdorff space, is a $T_1$ space in which a point outside a closed set can be separated from it by a continuous function: after rescaling, $f(x)=0$ and $f{|}_C=1$ with values in $[0,1]$.

The product topology has the subbasis of a topologyA map into the product is continuous exactly when each coordinate map is continuous. One direction follows by composition with the continuous product projections; the other follows by taking inverse images of the displayed subbasic sets.

Define the diagonal evaluation map $\Delta (x)=(f_{\lambda }(x){)}_{\lambda \in \Lambda }$. Its coordinates are continuous, so the map is continuous by continuity into a product space. Point separation makes it injective. For a basic open set $B=f_{\lambda }^{-1}(U)$,Every open subset of $X$ is a union of such $B$, so its image is open in $\Delta (X)$. The inverse on the image is continuous, and $\boxed{\Delta \text{ is a topological embedding}}$. If $X$ is $T_0$, any two distinct points are distinguished by an open set and then by one basic neighborhood within it. Equality of all coordinates would prevent that distinction. Thus the basis hypothesis alone implies point separation in a $T_0$ space. The unsubscripted $f$ in the printed separation hypothesis is understood as the chosen coordinate $f_{\lambda }$.

For the Sierpiński space $S=\{0,1\}$ with opens $\varnothing ,\{1\},S$, each open $U\subseteq X$ gives a continuous characteristic map ${\chi }_U:X\to S$ with ${\chi }_U^{-1}(\{1\})=U$. These inverse images form the whole topology, hence a basis. In a $T_0$ space they separate points, proving the embedding into $S^{\tau }$.

For a Tychonoff space, use all continuous maps $X\to [0,1]$. Given $x\in U$ with $U$ open, complete regularity supplies $f(x)=0$ and $f=1$ on $X\setminus U$. Then $f^{-1}([0,1/2))$ is an open neighborhood of $x$ contained in $U$. These inverse images form a basis. The diagonal theorem gives an embedding into a product of closed unit intervals.

A three-point space works even without any separation axiom. TakeFor each open $U$ of an arbitrary space, let $g_U=2$ on $U$ and $0$ elsewhere. It is continuous. For each point $a$, let $h_a=1$ at $a$ and $0$ elsewhere. It is also continuous, because its image lies in the subspace $\{0,1\}$ of $Y$, which has the indiscrete topology; equivalently $h_a^{-1}(\{2\})=\varnothing$. The $g_U$ supply every open set as an inverse image, and the $h_a$ distinguish every pair of points. Using the diagonal evaluation map for their combined family gives a universal three-point space for product embeddings:

Finally embed the given completely regular Hausdorff space into a cube $K=[0,1{]}^I$. The Tychonoff theorem makes $K$ compact; it is Hausdorff and hence normal. The image of the compact subset $C$ is compact, so closed in $K$. The Tietze extension theorem says that a continuous bounded real-valued function on a closed subset of a normal space extends continuously to the whole space. The given function on $C$ is bounded by compactness, so apply this theorem to its transported function on the image of $C$ in $K$. Restrict the extension to the image of $X$ and pull it back along the embedding. This yields $\boxed{F:X\to \mathbb{R}\text{ continuous},F{|}_C=f}$. This proves that compact subsets of Tychonoff spaces are C-embedded, using normality of the compact cube.

A family of open subsets is a subbasis of a topology when its finite intersections form a basis of a topology, including the empty intersection equal to the entire space. For the product topology, inverse images of open sets under the product projections form the standard subbasis of a topology. A map into the space is continuous if the inverse image of every subbasic set is open.