Cartesian product and closure Let $A \subset X$ and $B \subset Y$. Show that in the space $X \times Y$, $$\overline{A \times B}= \overline{A} \times \overline{B}$$
$\subset$: let $x \in \overline{A \times B}$.
Then for every open set $W$ containing $x$, $W \cap (A\times B) \neq \emptyset$
Since $W$ is an open set of $X \times Y$, $W= U \times V$ such that $U$ is an open set in $X$ and $V$ is an open set in $Y$.
We have: $W \cap (A \times B)\neq \emptyset \iff (U \times V) \cap (A\times B) \neq \emptyset \iff (U \cap A) \times (V \cap B) \neq \emptyset \iff U\cap A \neq\emptyset \text{ or } V \cap B\neq \emptyset$
Since $x \in W= U \times V$, $x= (u,v)$ such that $u \in U$ and $v \in V$
$\bullet$ There exists an open set $U$in $X$ containing $u$ such that $U \cap A= \emptyset$
Therefore $u \in \overline{A}$
$\bullet$ Similarly $v \in \overline{B}$
Since $x= (u,v) \in U \times V, x\in \overline{A} \times \overline{B}$
Hence: $\overline{A \times B} \subset \overline{A} \times \overline{B}$
$\supset$: Conversly, let $x \in \overline{A} \times \overline{B}$
$x=(a,b) \in \overline{A} \times \overline{B}$ such that $a\in \overline{A}$ and $b\in \overline{B}$
$\bullet$ There is an open set $U$ in $X$ containing $a$ such that $U \cap A\neq \emptyset$
$\bullet$ There is an open set $V$ in $Y$ containing $b$ such that $V \cap B\neq \emptyset$
Therefore $U \times V$ is an open set in $X \times Y$ containing $x=(a,b)$ such that $(U\times V) \cap (A\times B)\neq \emptyset \Rightarrow x\in \overline{A \times B}$
$\overline{A} \times \overline{B} \subset \overline{A \times B}$
Hence: $\overline{A \times B}= \overline{A} \times \overline{B}$
 A: A shorter route for one of the parts:
$A\times B\subset\overline{A}\times\overline{B}$ and $\overline{A}\times\overline{B}=\left(\overline{A}\times Y\right)\cap\left(X\times\overline{B}\right)$
is - as intersection of closed sets - closed in the producttopology.
Then $\overline{A\times B}\subset\overline{A}\times\overline{B}$.
addendum
The other part:
Let $\left(a,b\right)\in\overline{A}\times\overline{B}$ and let $W$
be open in $X\times Y$ with $\left(a,b\right)\in W$. Then some $U$ open
in $X$ and some $V$ open in $Y$ will exist with $\left(a,b\right)\in U\times V\subset W$.
Then $U\cap A\neq\emptyset$ and $V\cap B\neq\emptyset$. Let $x\in U\cap A$
and let $y\in V\cap B$. Then $\left(x,y\right)\in W\cap\left(A\times B\right)$.
This proves that every open set that contains $\left(a,b\right)$
has a non-empty intersection with $A\times B$. So $\left(a,b\right)\in\overline{A\times B}$.
A: Your proof for $\supseteq$ is correct, but your proof for $\subseteq$ isn't quite. You said

Since $W$ is an open set of $X \times Y$, $W= U \times V$ such that $U$ is an open set in $X$ and $V$ is an open set in $Y$.

This isn't true, since sets of the form $U \times V$ for $U \subseteq X$ and $V \subseteq Y$ open don't form a topology on $X \times Y$; but they do generate a topology. In other words, any open $W \ni x$ can be written as a union of sets of the form $U \times V$. This means that we can say $x \in U \times V$ for some $U$ open in $X$ and $V$ open in $Y$, such that $(U \times V) \cap (A \times B) \ne \varnothing$.
Most of your other reasoning goes through fine, except most (all?) of the times where you've written $= \varnothing$ you should have written $\ne \varnothing$.
