If $G\subset E, G$ is open and $E$ satisfies $\overline{E^0}\subset G$. Prove $G$ is closed too.
I'm relatively new to topology and the following is what I "THINK" I've proven but not exactly sure. I would really appreciate if someone corrected me if I'm wrong or if there is more direct way to do this as I don't feel too confident in my proof.
For reference these are the definitions I was given and therefore been using for the proof.
$x\in\overline{A}\iff\forall(G\in\tau, x\in G),(G\cap A\neq\emptyset=\exists y\in G\cap A)$.
$x\in A^0\iff\exists G\in\tau,x\in G\subset A$.
$x\in A^a\iff\forall(G\in\tau,x\in G),\exists y\neq x\in G\cap A$.
$\textbf{Proof:}$ Assume $G\subset E, G\in\tau$ and $E$ satisfies $\overline{E^0}\subset G$.
Consider $x\in\overline{E^0}$. Then $x\in G$.
Thus, $\exists y\in G\cap E^0$ and consequently $y\in G$ and $y\in E^0$.
Since $y\in E^0, \exists G\in\tau, y\in G\subset E$ and thus $y\in E$.
Then $y\in E$ and $y\in G\Rightarrow \exists y\neq x\in E\cap G$.
Hence $x\in G^a$.
Since $x\in G$ then $G^a\subset G$.
By definition, $G$ must be closed as well.
QED
I feel like this is totally wrong but at least I feel like I'm in the right track. I do know that a set is closed if it's complement is open but I didn't know how to formulate the proof in that manner. Any advice is greatly appreciated.
Since $G\subset E,$ then $G^\circ \subset E^\circ$ and thus $\overline{G^\circ}\subset \overline{E^\circ}\subset G.$ Since $G$ is open, $G=G^\circ,$ hence $\overline{G}\subset G.$ This shows $G=\overline{G}$ and therefore $G$ is closed.