I'm not sure about the proof I gave, But as I checked, it didn't use full AC, but as Asaf mentioned in a comment it uses DC. The following theorem is due to Pontrjagin. See Book by Montgomery and Zippin(Page 29).
I will give a sketch of proof.

**Note**: Maybe it's needed to add some separation axiom to the following theorem. please edit it, if needed.

### Theorem: Every $T_{0}$ topological group is completely regular.

*Proof.*
It's enough to prove that a given topological group $(G,*)$ is completely regular at $e$. Let $F$ be a closed set not containing $e$. Put $O=F^{c}$. Choose symmetric open neighborhoods of $e$, $U_{n}$ By continuity of $*$, such that $U_{0}=O$(w.l.o.g assume $O$ is symmetric) $U_{n+1}^2 \subseteq U_{n} \cap O ~~~~n=0,1,2...$.

Now for rational numbers of the form $r=\frac{k}{2^n}~~k \in \{1,2,3,...2^n \},~~ n \in \{0,1,2,...\}$ inductively define open neighborhood $V_{r}$ of $e$ such that:

*1)* $V_{\frac{1}{2^n}}=U_{n}~~~~~\forall n$

*2)* $V_{\frac{2k}{2^{n+1}}}=V_{\frac{k}{2^n}}$

*3)* $V_{\frac{2k+1}{2^{n+1}}}=V_{\frac{1}{2^{n+1}}}V_{\frac{k}{2^n}}$

The definition does not depend to the representation of $r$ and the family $V_{r}$ has the following properties:

*4)* $V_{\frac{1}{2^n}}V_{\frac{m}{2^{n}}} \subseteq V_{\frac{m+1}{2^n}}~~~~~~~~~m+1 \leq 2^n$

*5)* $V_{r} \subseteq V_{s}~~~~~$if $~r <s<1$
Now define the function $f:G \longrightarrow [0, 1]$ as follows:

$f(x)=0~$ if $x \in \bigcap \limits_{r} V_{r}$

$f(x)=1$ if $~~x \notin V_{1}$, and in other cases define:

$f(x)= \sup \{r \leq 1 : x \notin V_{r}\}$

### Claim: $f$ is as required.