= Solution
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 topology>
$$
\boxed{\{\pi_\lambda^{-1}(U):\lambda\in\Lambda,\ U\text{ open in }X_\lambda\}.}
$$
A 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> $B=f_\lambda^{-1}(U)$,
$$
\Delta(B)=\Delta(X)\cap\pi_\lambda^{-1}(U).
$$
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.
\b[A three-point space works even without any separation axiom.] Take
$$
\boxed{Y=\{0,1,2\},\qquad\tau_Y=\{\varnothing,\{2\},Y\}.}
$$
For 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>:
$$
\boxed{X\hookrightarrow Y^{\tau\sqcup X}.}
$$
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.
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