Let $S$ be a topological space. I need to prove that $S$ is Hausdorff if and only if the intersection of all closed neighborhoods of a point equals the point itself. I have a proof, given below, that I am hoping to get some feedback on.
Assume that $S$ is Hausdorff. Assume by way of contradiction that the intersection of all closed neighborhoods of $q\in S$ also contains a point $p\in S$ but for which $p\neq q$. Since $S$ is Hausdorff we know that there exist open neighborhoods $U_q$ of $q$ and $U_p$ of $p$ for which $U_q\cap U_p=\emptyset$. Now consider the set $S\setminus U_p$ and observe that $U_q\subset S\setminus U_p$. But $U_p$ is open so that $S\setminus U_p$ is a closed neighborhood of $q$. Hence there is a closed neighborhood of $q$ excluding $p$, a contradiction.
Now suppose that the intersection of all closed neighborhoods of a point equals the point itself. Let $q, p\in S$ with $q\neq p$. By assumption there is a closed neighborhood $V_q$ of $q$ that does not contain $p$. Since $V_q$ is closed $S\setminus V_q$ is an open set containing $p$; therefore, $S\setminus V_q$ is an open neighborhood of $p$. Hence, we have neighborhoods $V_q$ of $q$ and $S\setminus V_q$ of $p$ that are disjoint. Hence the space is Hausdorff.