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.

Since $M$ is a metric space, we can argue using sequences:
Suppose $A$ is open, $X\subseteq M$, and let $a\in A\cap  \overline{X}$. Then  since $a\in \overline{X}$, there is a sequence $(x_n)$ in $X$ with $x_n\rightarrow a$. But, as $a\in A$ and  $A$ is open, $(x_n)$ is eventually in $A$. Thus, there is a sequence (a tail of $(x_n)$) in $A\cap X$ that converges to $a$. Thus $a\in\overline{A\cap X}$. Since $a$ was an arbitrary element of $A\cap \overline X$ we have $A\cap \overline{X}\subseteq\overline{A\cap X}$.
Now suppose that $A$ is not open. Then there is an $a\in A$ and a sequence $(x_n)$ converging to $a$ with $x_n\in A^C$ for each $n$ ($A$ is open iff whenever $z_n\rightarrow z\in A$, then $(z_n)$ is eventually in $A$). Set $X=\{\,x_1,x_2,\ldots\,\}$. Then $a\in A\cap \overline{X}$.  But $A\cap X=\emptyset$; and so $a\notin \overline {A\cap X}$. Thus $A\cap \overline{X}\not\subseteq\overline{A\cap X}$.

The arguments in the other answers are better of course, as they show the result holds for any topological space.
I'm not sure that $M$ being metric would lead to a simpler proof...
