Every point closed $\stackrel{?}{\Rightarrow}$ space is Hausdorff If a topological space is Hausdorff, then every point is closed.
Is the converse true?
Edited: Let $G$ be a topological group and $H$ the intersection of all
neighborhoods of zero. Since every coset of $H$ is closed, every point
of $G/H$ will be closed. Why does that make $G/H$ Hausdorff?

Notice that $H$ is a closed normal subgroup of $G$, for that see e.g. this for proof that it is a closed subgroup (equal to $\operatorname{cl} \{e\}$), and for normality just notice that conjugation preserves the neighbourhoods of identity (as a set), so it does preserve intersection as well.
From that we see that $G/H$ is a topological group.
It is a known fact that for topological groups, $T_0$ implies completely regular Hausdorff. Every point being closed is equivalent to $T_1$, from which $T_{3\frac {1}{2}}$, so in particular $T_2$, follows.
A proof can be found in many places, e.g. Engelking's General Topology iirc.
A short one for closed $\{e\}\implies T_2$: notice that Hausdorffness is equivalent to the diagonal being closed. But the diagonal is the preimage of identity by the map $(x,y)\mapsto xy^{-1}$.
In general, we do not have the implication, as shown by e.g. the cofinite topology on an infinite space.
