How I can prove that $\overline{A \times B}=\overline{A} \times \overline{B}$?
I start proving that $\overline{A} \times \overline{B} \subseteq \overline{A \times B}$.
Taking $(x,y) \in \overline{A} \times \overline{B},\ \ \ \forall U \times V \in \mathscr{N}_{(x,y)}$,
$x \in \overline{A} \Rightarrow U\cap A \not= \emptyset \ $ and $\ y \in \overline{B} \Rightarrow V\cap B\not= \emptyset \ $ so
$U \times V \cap A \times B \not= \emptyset \Rightarrow (x,y) \in \overline{A \times B}$
Is this part correct? And how can I prove the other part?
That part is correct, but you could use more words and fewer symbols: let $(x,y) \in \overline{A} \times \overline{B}$. To see that $(x,y) \in \overline{A \times B}$, we take a basic open open neighbourhood of $(x,y)$, so one of the form $U \times V$ where $U$ is an open neighbourhood of $x$ and $V$ an open neighbourhood of $y$ and show it intersects $A \times B$: as $U$ is an open neighbourhood of $x$ and $x \in \overline{A}$ we know that there is some $a \in A \cap U$ and similarly, as $y \in \overline{B}$ we have some $b \in B \cap V$, but then $(a,b) \in (A \times B) \cap (U \times V) \neq \emptyset$ and we're done.
For the reverse, we take $(x,y) \in \overline{A \times B}$ and we will show that $x \in \overline{A}$ and $y \in \overline{B}$, so that $(x,y) \in \overline{A} \times \overline{B}$, showing the inclusion.
To see that $x \in \overline{A}$: let $U$ be an open neighbourhood of $x$. Then $U \times Y$ is an open neighbourhood of $(x,y)$ so the assumption that $(x,y) \in \overline{A \times B}$ gives us $(a,b) \in (U \times Y) \cap (A \times B)$. In particular we have that $a \in U \cap A \neq \emptyset$. As $U$ was arbitrary, we have that $x \in \overline{A}$ as claimed. To show that $y \in \overline{B}$ is similar and left to the reader, using sets of the form $X \times V$ instead.
As an alternative to that last proof, note that $$\overline{A} \times \overline{B} = \pi_X^{-1}[\overline{A}] \cap \pi_Y^{-1}[\overline{B}]$$ where $\pi_X$ and $\pi_Y$ are the continuous projections so that $\overline{A} \times \overline{B}$ is closed in $X \times Y$. It obviously contains $A \times B$ so also $\overline{A \times B}$, by minimality of the closure.