$G_n$ is a countable family of dense open sets
X is a metric space
I would like to show that if $\bigcap G_n$ is not dense in X then $int(\overline{\bigcap G_n})$ is also not dense in X.
I have tried the following:
Since $\bigcap G_n$ is not dense, $\overline{\bigcap G_n} \neq X$ thus $\exists x \in X$ st $ x \notin \overline{\bigcap G_n}$. Hence we have that $x \notin \bigcap G_n$ and x is not a limit point of $\bigcap G_n$.
Since $ int(\overline{\bigcap G_n}) \subset \overline{\bigcap G_n}$ we see $ x \notin int(\overline{\bigcap G_n})$ and that x is not a limit point of $int(\overline{\bigcap G_n})$. Which immplies $ x \notin \overline{int(\overline{\bigcap G_n})}$ so $int(\overline{\bigcap G_n}) \neq X$ therefore $int(\overline{\bigcap G_n})$ is not dense in X.
Does this look fine? If not, how would i correct go about correcting it.
Your proof looks fine to me. In fact, if we put $A = \bigcap G_n$, what we have to prove is that, if $A$ is not dense in $X$ (i.e,. $\overline{A} \neq X$), then $\mathrm{int}(\overline{A})$ is not dense in $X$ (i.e., $\overline{\mathrm{int}(\overline{A})} \neq X$), and you have proved that. This could be proved more tersely by observing that if $\overline{A} \neq X$, then, because $\mathrm{int}(\overline{A}) \subseteq \overline{A}$, we have $\overline{\mathrm{int}(\overline{A})} \subseteq \overline{\overline{A}} = \overline{A} \neq X$. However, terseness is not always good: I've implicitly used the facts that for any $Y \subseteq X$, $\mathrm{int}(Y)\subseteq Y$ and $\overline{\overline{Y}} = \overline{Y}$.