= Connected complement of a product of proper subsets
{title2=$(X\times Y)\setminus(A\times B)$}
Suppose $X,Y$ are nonempty <connected spaces>, and $A\subsetneq X$, $B\subsetneq Y$ are proper <subsets>. Then $C=(X\times Y)\setminus(A\times B)$, with its <subspace topology> from the <product topology>, is a <connected space>. No <closed set> or <open set> hypothesis on $A,B$ is needed.
Choose $x_0\in X\setminus A$, $y_0\in Y\setminus B$. The two <topological subspaces> $\{x_0\}\times Y$ and $X\times\{y_0\}$ are <homeomorphic> to the factors, intersect at $(x_0,y_0)$, and lie in $C$. Their <union> is a <connected subset> $H$. Every $(x,y)\in C$ has $x\notin A$ or $y\notin B$. In the first case its entire vertical copy $\{x\}\times Y$ lies in $C$ and meets $H$; in the second case the horizontal copy $X\times\{y\}$ does so. Thus $C$ is a <connected union with a connected hub>.
Both properness assumptions are needed for this general result. Taking $X=Y=\mathbb R$, $A=X$ and $B=\{0\}$ gives the two disjoint <open> half-planes $y<0$ and $y>0$, a <separation of a topological space> of the complement.
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