= Solution
Choose $x_0\in X\setminus A$ and $y_0\in Y\setminus B$, possible because the two subsets are proper. The cross
$$
H=(X\times\{y_0\})\cup(\{x_0\}\times Y)
$$
lies 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 gives
$$
\boxed{(X\times Y)\setminus(A\times B)\text{ is connected}.}
$$
Neither proper subset needs to be <open>, <closed>, or <connected>.
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