I want to show that $G$ is open in $X$ iff $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$.
This direction $\rightarrow$ is easy.
I'm stuck in showing that $\overline{G \cap \overline{A}} = \overline{G \cap A}$ for every $A \subset X$ implies $G$ is open in $X$.
So I guess the game is to find suitable A and apply the hypothesis, I tried taking A as the interior of G and also as the interior of the complement of G but didn't work (at least I didn't see it).
Any ideas?