Closure of product 
Let $A\subseteq X$ and $B\subseteq Y$. Show that in the space $X\times Y$, $\overline{A\times B}=\overline{A}\times\overline{B}$.

Let $(x,y)\in\overline{A\times B}$. Then every open set in $X\times Y$ containing $(x,y)$ also contains an element in $A\times B$. We will prove that $x\in\overline{A}$. Take an open set $U$ containing $x$. Then $U\times Y$ is an open set containing $(x,y)$. So it contains an element in $A\times B$. So $U$ contains an element of $A$. We can prove similarly that $y\in\overline{B}$.
Now let $(x,y)\in\overline{A}\times\overline{B}$. Then every open set in $X$ containing $x$ also contains an element of $A$. Every open set in $Y$ containing $y$ also contains an element of $B$. Take an open set $U$ in $X\times Y$ containing $(x,y)$. What can we do?

Every neighborhood of $(x,y)$ contains a product $U\times V$, where $U$ and $V$ are open neighborhoods of $x$ and $y,$ respectively. Now there is an $a\in U\cap A$ and a $b\in V\cap B.$ Can you finish the proof?
