 Solution

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.