Say we have $X$ a $T^1$ space, and $A$ a subset of $X$ We aim to show that the intersection of all open sets containing $A$ equals $A$ By de Morgan's laws, that would be true if the complement of $A$ $A^c$ equalled the union of all closed sets in $A^c$ Let's call this union of closed sets $C$

Each set that makes up $C$ is contained by $A^c$ so $C\subset A^c$ If we could show $A^c\subset C$ we'd be done.

Since $X$ is $T^1$ each singleton in $A^c$ is closed. Their union, a subset of $C$ contains $A^c$ so we're through.

Now suppose we know that in some topological space $X$ any subset $A$ of $X$ is the intersection of all open sets containing $A$ Given $x\neq y$ we're looking for an open set containing $x$ but not $y$ to show that $X$ is $T^1$

$$\{x\}=\bigcap_{\substack{U\text{ open}\\U\ni x}}U$$

by hypothesis. If all open sets containing $x$ contained $y$ $y$ would be in the intersection; since $y$ isn't in the intersection, $X$ must be $T^1$