You could find a routin proof in the book "Topological Ring" written by Seth Warner. In this book at page 21 in Theorem 3.4 you could see the following Proposition:

Theorem: Let $G$ be a topological group. The following statements are equivalent:

- {$0$} is closed.
- {$0$} is an intersection of the neighborhoods of zero.
- $G$ is Hausdorff.
- $G$ is regular.

You could also find the improvement of it in the book "**Topology for analysis**" Written by **Albert Wilansky**. In section 12 at page 243 You could see the following theorem:

THEQREM: Every topological group is completely regular. The following conditions on
a topological group $G$ are equivalent:

- $G$ is a $T_0$ space.
- $G$ is a Tychonoff space.-
- $\cap${$U:U$ is a is a neighborhood of $e$}={$e$}

The Proof of Complete regularity Has more details. I think The proof is in the level of Urysohn Lemma.

But if you are interested in the general case, I Suggest You look at the section "uniformity" which were discussed in the following books:

- Topology for Analysis: Chapter 11
- General topology: Stephen Willard: chapter 9