$\overline{U}\cap V\subseteq\overline{U\cap V}$ In particular if the equality is generally false, is it true if $V$ is open?
Could someone help, me please? 

If $V$ is open, this holds: let $x \in \overline{U} \cap V$, so $x \in V$ and $x \in \overline{U}$. 
Then let $O$ be an arbitrary open subset containing $x$. Then $x \in O \cap V$ and this set is also open, so it intersects $U$ as $x \in \overline{U}$. So $$\emptyset \neq (V \cap O) \cap U = O\cap (U \cap V)$$ for any open neighbourhood $O$ of $x$, which implies $x \in \overline{U \cap V}$, as required.
