In a topological space $X, G$ is open iff $G\cap\overline{A}\subseteq \overline{G\cap A}$ 
In a topological space $X$, $G$ is open iff $G\cap\overline{A}\subseteq \overline{G\cap A}$, where
  $A$ is an arbitrary set in $X$.

To prove this I started from definition of an open set, I could not complete it.
I even tried to prove it in metric space using def I could not complete it
 A: Take $A=X-G$. Then $G\cap A=\varnothing$, so $\overline{G\cap A}=\varnothing$. Therefore, you must have $G\cap\overline{A}=\varnothing$. This means that $\overline{X-G} = \overline{A}\subseteq X-G$, hence $\overline{X-G}=X-G$. That is, the complement of $G$ is closed.
Conversely, suppose that $G$ is open, and let $A$ be arbitrary. Let $x\in G\cap \overline{A}$. To show that $x\in \overline{G\cap A}$, we just need to show that any open set that contains $x$ will intersect $G\cap A$ nontrivially. Let $\mathcal{O}$ be an open set containing $x$. Then $\mathcal{O}\cap G$ is an open set containing $x$ (since $x\in G$), so $\mathcal{O}\cap G\cap A\neq\varnothing$ (since $x\in\overline{A}$). Thus, $\mathcal{O}\cap (G\cap A)\neq\varnothing$. This proves that $x\in\overline{G\cap A}$, giving the desired inclusion. 
A: Assume that $G \cap \overline{A} \subset \overline{G\cap A} $ then
\begin{equation}\label{1}
\begin{aligned}
G \cap \overline{A} \subset \overline{G\cap A} & \Longleftrightarrow X \backslash (\overline{G\cap A}) \subset X \backslash ( G \cap \overline{A}) \Longleftrightarrow X \backslash int(G \cap A) \subset X \backslash ( G \cap \overline{A})
\end{aligned}
\end{equation}
Take $A = X$ then 
\begin{equation}
X \backslash int(G) = X \backslash int(G \cap X) \subset X \backslash G \cap \overline{X} =  X \backslash G ,
\end{equation}
 i. e., $G \subset int(G)$ and therefore $G$ is open.
