Show that G is open in X iff $\overline{G \cap \bar{A}}=\overline{G \cap A}$ for all $A\subset X$

Question

I have tried

Suppose $G$ is open , we claim that $\overline{G \cap \bar{A}}=\overline{G \cap A}$ for all $A\subset X$

Since $G \cap A \subseteq G \cap \overline A$

$\Rightarrow \overline{G\cap A} \subseteq \overline{G\cap \overline A}$

Let $x \in \overline{G\cap \overline A}$ and $S_r(x)$ be a $nbd$ of $x$, then $S_r(x) \cap (G \cap \overline A) \neq \phi$, then $\exists$ $y \in S_r(x) \cap (G \cap \overline A) $ there exists $r'$ such that $S_{r'}(y) \subset (S_r(x)\cap G)$ and $S_{r'}(y) \cap A \neq \phi$ $\Rightarrow S_r(x) \cap (G\cap A) \neq \phi$

$\therefore$ $x \in \overline{G\cap A}$

Thus $\overline{G \cap \bar{A}}=\overline{G \cap A}$ for all $A\subset X$

Converse :

if $\overline{G \cap \bar{A}}=\overline{G \cap A}$ for all $A\subset X$, then $G$ is open.

Please tell how to show that $G$ is open. Thank you

Answer 1

Substitute $A=G^c$. Then,

$$\overline{G\cap \overline{G^c}}=\overline{G\cap G^c}=\varnothing \Rightarrow G\cap \overline{G^c}=\varnothing \Rightarrow \overline{G^c} =\overline{X\setminus G}\subset X\setminus G=G^c.$$

Therefore $G^c$ is closed and hence $G$ is open.

Answer 2

Hint: take $A=X\setminus G$. What does the condition tell you?