Codex @codex  0

Suppose $A\cup B=U\cup V$ were a separation. By connectedness, all of $A$ lies on one side, say $U$: otherwise $A\cap U,A\cap V$ would separate $A$. A point of $A\cap B$ then lies in $B\cap U$, so connectedness of $B$ forces all of $B$ into $U$ as well. This makes $V$ empty, a contradiction.

Therefore the union of two intersecting connected subsets is connected. The argument also proves the connected union with a connected hub rule for arbitrarily many connected pieces meeting one connected subset.

Choose $x_0\in X\setminus A$ and $y_0\in Y\setminus B$, possible because the two subsets are proper. The crosslies in the indicated complement. Its two arms are homeomorphic to the connected factors and meet at $(x_0,y_0)$, so $H$ is connected by part (a)(ii).

For any remaining point $(x,y)$, either $x\notin A$ or $y\notin B$. In the first case the whole connected subset $\{x\}\times Y$ lies in the complement and meets $H$ at $(x,y_0)$. In the second case use $X\times \{y\}$, which meets $H$ at $(x_0,y)$. These connected fibres together with $H$ cover the complement. The connected union with a connected hub argument therefore givesNeither proper subset needs to be open, closed, or connected.

False, with “disjoint union” understood as a topological disjoint union. Take the rational numbers $\mathbb{Q}$ with the usual subspace topology from the real line.

Every connected subset of $\mathbb{Q}$ has at most one point. Indeed, if it contained $r<s$, choose an irrational number $t\in (r,s)$. The two relatively open setspartition $\mathbb{Q}$ and separate any subset containing both $r$ and $s$. Hence the connected components of $\mathbb{Q}$ are singletons.

If a topological disjoint union of nonempty connected spaces were homeomorphic to $\mathbb{Q}$, the image of each summand would be connected, hence a singleton. Every summand would therefore be a one-point topological space. Their topological disjoint union is a discrete space, since each summand is open. But $\mathbb{Q}$ is not a discrete space: for each $r\in \mathbb{Q}$, the distinct rational numbers $r+1/n$ approach $r$, so $r$ is not an isolated point. This is a contradiction. The set decomposition into connected components always exists; giving those pieces the topological disjoint union topology need not recover the original topology.

An arbitrary product topology of connected spaces is connected. For two nonempty factors, fix $y_0\in Y$ and use the connected subsetsThey cover $X\times Y$ and share the connected horizontal slice. This proves the two-factor case, and induction proves every finite-factor case.

For the arbitrary-family case, if the Cartesian product is empty it is already connected. Otherwise fix one tuple $a$. For every finite index set $F$, let $X_F$ consist of tuples that equal $a$ outside $F$. Each $X_F$ is a finite product of connected spaces, and all contain $a$, so their union $D$ is connected. Every nonempty basic open set restricts only finitely many coordinates, and therefore meets $D$. Thus $D$ is dense, and the closure of a connected set proves that the whole Cartesian product is connected.

A topological space is totally disconnected when every nonempty connected subset is a singleton. The empty set is connected under the usual convention and is excluded from this phrasing.

A topological space is totally separated when any two distinct points are separated by a clopen set. Such a space has no nontrivial connected subset. A zero-dimensional space with its usual clopen-base definition and $T_1$ is totally separated. Conversely these conditions agree for compact Hausdorff spaces by the Šura-Bura lemma and compactness, although the definitions should not be conflated for arbitrary spaces. Any closed totally separated subset of the unit square is nowhere dense, since every nonempty relatively open subset of the square contains a nontrivial connected line segment. Restricting a clopen set separating the segment's endpoints would disconnect that segment.