I am sure this is really a trivial result, but I would like to check my proving skills (plus I find it is sort of related to nowhere dense sets, that are quite difficult for me to digest.
Proposition:
Let $(X, \tau)$ be a topological space. If $A \in \tau$, then int$(\bar{A}) = A$.
Proof:
(only if)
Assume that $x \notin A$, and let $G \in \tau$ be arbitrary and such that $G \subseteq \bar{A}$. Assume by contradiction that $x \in G$. Then, we reach a contradiction because $x \in G \subseteq \bar{A}$ with $G \in \tau$, and $x \notin A$, where $A$ is the largest open subset of $\bar{A}$.(if) Assume that $x \notin \text{int}(\bar{A})$. Hence, for all $G \in \tau$, if $G \subseteq \bar{A}$, then $x \notin G$. Thus, $A \subseteq \bar{A}$ implies $x \notin A$ and the result is established.
Of course, as always any feedback is most welcome.
Thank you for your time.