I know that this question is already answered in the site but I did it in a different way but I don't know if it is correct.
My attempt:
Suppose that $(X\times Y) \setminus (A\times B)$ is disconnected.
Then there exist non-trivial (I know this because of $A \neq X$ and $B \neq Y$) disjoint open sets $U$ and $V$ such that $U \cup V = (X\times Y) \setminus (A\times B)$.
Then we know that $(U \cup V) \cup A\times B = X\times Y$.
Let's call $C=U \cup V$ and $D=A \times B$ and by construction these are disjoint.
Then $C \cup D = X \times Y$ but that is a contradiction because $X \times Y$ is connected.
So $(X\times Y) \setminus (A\times B)$ is connected.
Best Answer
This does not work. $U$ and $V$ are open in $X\times Y \setminus A \times B$, but for a decomposition as you want you need $U\cup V$ to be open in $X \times Y$. And you also need $A\times B$ to be open in $X \times Y$, which can only happen if $A$ and $B$ are. And you are only allowed to use that $A$ and $B$ are proper subsets. You assume way too much, and even inconsistent things.