Question about closure of the product of two sets

$(\subseteq)$: The product of closed sets $\overline{A} \times \overline{B}$ is closed. For every closed $C$ that contains $\overline{A} \times \overline{B}$, $A \times B \subseteq C$ so $\overline{A \times B} \subseteq \overline{\overline{A} \times \overline{B}} = \overline{A} \times \overline{B}$.

$(\supseteq)$: Choose any $(a,b) \in \overline{A} \times \overline{B}$. Notice that for every open neighborhood $W \subseteq X \times Y$ that contains $(a, b)$, $U \times V \subseteq W$ (by the definition of the product topology) for some open neighborhood $U$ of $a$ and some open neighborhood $V$ of $b$. By the definition of closure points, $U$ intersects $A$ at some $a'$. Similarly, define $b' \in V \cap B$. Hence, $(a', b') \in W \cap (A \times B)$.

To summarize, every open neighborhood $W \subseteq X \times Y$ that contains $(a,b)$ must intersect $A \times B$, therefore $(a,b) \in \overline{A\times B}$.

A set in $X\times Y$ is open iff it is the union of sets of type $U\times V$ with $U,V$ open.

Therefore, if $(x,y)\notin\overline{A\times B}$, there is $U\times V$ containing it and disjoint to $A\times B$. As $(U\cap A)\times (V\cap B)\subseteq (U\times V)\cap (A\times B)$, one of the sets $U\cap A$ or $V\cap B$ is empty, i.e. $x\notin \overline A$ or $y\notin \overline B$. This shows $\overline A\times\overline B \subseteq \overline{A\times B}$. The converse is not much different.