Theorem: Given a topological space with $A\subseteq X$, then $x\in \overline A$ iff for each neibourhood of $x$, $U\bigcap A\neq \emptyset$.
Proof:
$\rightarrow$ : Let $x\in \overline A$ and let $U$ be a neibourhood of $x$; then there exists open $G$ with $x\in G \subseteq U$. If $U\bigcap A=\emptyset$, then $G\bigcap A=\emptyset$, and so $A\subset X$ \ $G$ $\rightarrow$ $\overline A\subseteq X$ \ $G$ whence $x\in X$ \ $G$, thereby contradicting the assumption that $U\bigcap A=\emptyset$.
$\leftarrow :$ If $x\in X$\ $\overline A$ is an open nhd of $x$ so that, by hypothesis, $(X$\ $\overline A)\bigcap A\neq \emptyset$, which is a contradiction.
I lost the understanding at $\overline A\subseteq X$\ $G$. Why does it hold when $A\subseteq X$\ $G$?