Prove that G is open in space
Take $\displaystyle x\in\overline{G\cap\overline A}$ and let $\displaystyle V$ a neighborhood of $\displaystyle x$. We have $\displaystyle G\cap \overline A\cap V\neq \emptyset$. Let $\displaystyle a \in G\cap \overline A\cap V$. Use the fact that $\displaystyle G\cap V$ is a neighborhood of $\displaystyle a$.