Let $X$ be a topological space such that for any subset $S$, the closure $\bar{S}=\bigcap_{U\supset S} U$ over $U$ opens. Is $X$ normal?

Question

The converse is clearly seen to be true, but I don't see any particular way to prove this direction.

Answer 1

Your condition implies that every open set is closed, since if $S$ is open then $\bigcap_{U\supseteq S}U=S$ clearly (you can take $U=S$). Taking complements, this means closed sets are also open. Conversely, if open and closed sets are the same, clearly your condition holds, since by definition the closure of $S$ is the intersection of all closed sets containing $S$.

To answer your question, this condition clearly implies $X$ is normal if your definition of normal does not include points being closed (given two disjoint closed sets, they are also open and thus are separated by open sets). If your definition of normal does include points being closed, the indiscrete topology on any set with more than one point is a counterexample.

(In fact, with a bit more work you can show your condition is equivalent to the $T_0$ quotient of $X$ being discrete.)