How to show that topological groups are automatically Hausdorff?

A space $X$ is Hausdorff if and only if the diagonal $\Delta_X\subseteq X\times X$ is closed. Consider the map $G\times G\rightarrow G$ given by $(x,y)\mapsto xy^{-1}$. It is continuous by the axioms for a topological group, and the diagonal is the inverse image of the the identity $\{e\}$, which is closed by assumption. So $G$ is Hausdorff if $\{e\}$ is closed, i.e., if $G$ is $T_1$ (by homogeneity, $T_1$ for $G$ is equivalent to $\{e\}$ being closed).

Given $x\neq y$ in a Hausdorff $G$, let $U_x$ and $U_y$ be disjoint opens around $x$ and $y$, respectively. Both $U_xx^{-1}$ and $U_yy^{-1}$ are opens around $e$, so we can find open $V$ with $e\in V\subseteq U_xx^{-1}\cap U_yy^{-1}$. Then $Vx$ and $Vy$ are disjoint neighborhoods of $x$ and $y$, as desired.

A topological space $G$ is hausdorff iff the diagonal in $G\times G$ is closed. Can you see how the diagonal is the inverse image of a closed set under a continuous map?

By the above argument, there are two cases:

1. If $\{1\}$ is closed then the diagonal is closed (as it is the inverse image of $\{1\}$ under the continuous map $g_1 \times g_2\mapsto g_1g_2^{-1}$). So $G$ is Hausdorff.
2. If $\{1\}$ is open then $G$ is discrete, so it's Hausdorff.