Proposition 1   [1,2] Suppose $X$ is a topological space. Then $X$ is a $T_2$ space if and only if for all $x\in X$ , we have \begin{eqnarray} \label{ceq} \{x\} &=& \bigcap \{A \mid A\subseteq X\ \mbox{closed}, \mbox{$\exists$ open set}\ U\ \mbox{such that}\ x\in U\subseteq A\}. \end{eqnarray}

Proof. By manipulating the definition using de Morgan's laws, the claim can be rewritten as $$ \{x\}^\complement = \bigcup \{V \mid V\subseteq X\ \mbox{open}, \mbox{$\exists$ open set}\ U\ \mbox{such that}\ x\in U\subseteq V^\complement\}. $$ Suppose $y\in \{x\}^\complement$ . As $X$ is a $T_2$ space, there are open sets $U,V$ such that $x\in U, y\in V$ , and $U\cap V=\emptyset$ . Thus, the inclusion from left to right holds. On the other hand, suppose $y\in V$ for some open $V$ such that $\{x\}\subseteq V^\complement$ . Then $$ y\in V\subseteq \{x\}^\complement $$ and the claim follows.