[Proof Verification]: The closure of a set is closed.

Your proof is correct, maybe we can make it slightly faster.

Let $z$ be a limit point of $\overline A$. Every open set $U$ containing $z$ must contain a point $x$ in $\overline A$, if the point is in $A^\circ$ then $U$ must intersect $A$ because it contains a limit point of $A$. if $x$ is in $A$ then $U$ also intersects $A$.

So every limit point of $\overline A$ is a limit point of $A$, and $\overline A$ contains all of its limit points.