A characterization of open sets 
Let $(M,d)$ be a metric space. Then a set $A\subset M$ is open if, and only if, $A \cap \overline X \subset \overline {A \cap X}$ for every $X\subset M$.

This is a problem from metric spaces, but I think that it only requires topology. I don't know how to do it.

Suppose $A$ is open, let $X\subseteq M$, and let $a\notin \overline{A\cap X}$. Therefore, there exists an open set $U$ such that $x\in U$ and $U\cap (A\cap X)=\varnothing$. If $a\in A$, then $(U\cap A)$ is an open set that contains $a$, and whose intersection with $X$ is empty, so $a\notin \overline{X}$, hence $a\notin A\cap\overline{X}$. If $a\notin A$, then $a\notin A\cap\overline{X}$. Thus, we have shown that 
$$M-\overline{A\cap X} \subseteq M-(A\cap \overline{X}).$$ Taking complements proves the desired inclusion.
Conversely, suppose that for every $X\subseteq M$, we have $A\cap\overline{X}\subseteq \overline{A\cap X}$. We wish to show that $A$ is open. Let $X=M-A$. Then $A\cap\overline{X} \subseteq \overline{A\cap X} = \overline{A\cap(M-A)}=\varnothing$, so $\overline{X}\subseteq M-A = X$. Therefore, $X$ is closed.
