Proving that $U \in T$ has non-empty intersection with $A$ iff $U$ has non-empty intersection with $\overline{A}$ In a Hausdorff topological space, $(X, T)$, with non-empty subset $A \subset X$, how can we prove that an open set $U \in T$ has non-empty intersection with $A$ iff $U$ has non-empty intersection with $\overline{A}$? The $\Leftarrow$ direction is obvious, but I'd be interested in seeing how we can prove the opposite relation holds.

This pretty much follows straight from the definition of closure.

Let's say $U$ and $\overline{A}$ meet at some point $x$. If $U$ does not intersect $A$, then $\overline{A} - U$ is a closed set containing $A$ but not containing $x$. This is a contraction from the definition of closure, so $U$ intersects $A$ as desired.