Under certain conditions, is it true that $\overline{Z \cap U} = Z$? 
Let $U$ be an open of a topological space $X$ and let $Z$ be an irreducible closed in $X$. We have that $\overline{Z \cap U}$ is an irreducible and closed of $U$. Is it true that $\overline{Z \cap U} = Z$?

This inclusion I managed to prove:
$(\subset)$ Clearly $Z \cap U \subset Z$. Then we have that $\overline{Z \cap U} \subset \overline{Z}=Z$, since $Z$ is closed in $X$.
What mechanism can I use to prove $(\supset)$?
 A: If $Z\ne\emptyset$, then a necessary condition is obviously that $Z\cap U\ne\emptyset$.
You can even prove more:

if $Z$ is an irreducible closed set and $U$ is an open set such that $Z\cap U\ne\emptyset$, then $Z\subseteq U$.

Note that $F=X\setminus U$ is closed as well as $Z\cap F$, and
$$
Z=(Z\cap U)\cup(Z\cap F)\subseteq(\overline{Z\cap U})\cup(Z\cap F)
\subseteq Z
$$
Then $Z=(\overline{Z\cap U})\cup(Z\cap F)$. Since $Z$ is irreducible and $\overline{Z\cap U}\ne\emptyset$, we get that $Z\cap F=\emptyset$, so $Z\subseteq U$.
A: If $Z \cap U$ is nonempty, the equality $\overline{Z \cap U} = Z$ holds:
Suppose that $Z \cap U$ is nonempty. Clearly $Z \cap U \subset Z$. Since $Z \cap U$ is open in $Z$ and $Z$ is irreducible, we have that $Z \cap U$ is dense in $Z$, which implies that $\overline{Z \cap U} = Z$.
If $Z \cap U$ is empty, the equality holds only if $Z$ is empty, since $\overline{Z \cap U} = \emptyset = Z$. Otherwise, we have the following counterexample: Let $Z$ be non empty and $U=X \setminus Z$. Clearly $U$ is open in $X$, since $Z$ is closed in $X$. We have that $Z \cap U = Z \cap (X \setminus Z) = \emptyset$. Then $\overline{Z \cap U} = \emptyset \neq Z$.
