How to show that topological groups are automatically Hausdorff? On page 146, James Munkres' textbook Topology(2ed),

Show that $G$ (a topological group) is Hausdorff. In fact, show that if $x \neq y$, there is a neighborhood $V$ of $e$ such that $V \cdot x$ and $V \cdot y$ are disjoint.

Noticeably, the definition of topological group in Munkres's textbook differs from that in wikipedia.

A topological group $G$ is a group that is also a topological space satisfying the $T_1$ axiom, such that the map of $G \times G$ into $G$ sending $x \times y$ into $x \cdot y$ and the map of $G$ into $G$ sending $x$ into $x^{-1}$, are continuous maps.

 A: 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?
A: By the above argument, there are two cases: 


*

*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. 

*If $\{1\}$ is open then $G$ is discrete, so it's Hausdorff.

A: 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. 
