a space is T1 if and only if every subset A is the intersection of all open sets containing A

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 (http://planetmath.org/ASpaceIsT1IfAndOnlyIfEverySingletonIsClosed). 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 $x\ne y$, we’re looking for an open set containing $x$ but not $y$, to show that $X$ is $T^1$.

$$\{x\}=\bigcap _{\begin{array}{c}U\text{ open}\\ U\ni x\end{array}}U$$

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 $T^1$.