Hint:
a. For an arbitrary nbhood $W$ of $e$, let $U=W\cap W^{-1}$, then plainly $U=U^{-1}$, $U$ is a nbhood of $e$, and $U\subset W$. (now use,
theorem:
Neighborhood Axioms of $e$).
b.
If $G$ is a topologial group the following conditions are equivalent:
i) $G$ is a $T_0$ space.
ii) $G$ is a $T_1$ space.
iii) $G$
is a $T_2$
space.
iv) If $\beta_e$
is a fundamental system of neighborhoods of $e$
then
$\bigcap
β e ={e}$.
v) $\{e\}$ is a closed subgroup of $G$.
c. Every topological group is regular.
I'll refer you to a few books on topological groups. Hopefully useful to you.
A. Arhangel'skii, M. Tkachenko, Topological Groups and Related Structures.
T. Husain, Introduction to topological groups.
E. Hewitt, K. A. Ross, Abstract Harmonic Analysis: Volume 1.
In fact, you don't quite have to assume that $A$ and $B$ are disjoint closed sets, or even that they have disjoint closures; it's enough to assume that they are separated sets, i.e., each is disjoint from the closure of the other, a condition which is plainly necessary as well as sufficient.
Theorem. Let $A$ and $B$ be subsets of a metric space $(X,d)$. If $A\cap\overline B=\emptyset$ and $B\cap\overline A=\emptyset$, then there are disjoint open sets $U$ and $V$ such that $A\subseteq U$ and $B\subseteq V$.
Proof. If one of the sets is empty, it's easy; e.g., if $A=\emptyset$, we can take $U=\emptyset$ and $V=X$. So we may assume that $A$ and $B$ are nonempty sets. In that case the functions
$$x\mapsto d(x,A)=\inf\{d(x,a):a\in A\}$$
and
$$x\mapsto d(x,B)=\inf\{d(x,b):b\in B\}$$
are well-defined and continuous. It follows that the sets
$$U=\{x:d(x,A)\lt d(x,B)\}$$
and
$$V=\{x:d(x,B)\lt d(x,A)\}$$
are open sets, and of course they are disjoint. Finally, we have $A\subseteq U$ because $d(x,A)=0\lt d(x,B)$ for all $x\in A$, and $B\subseteq V$ because $d(x,B)=0\lt d(x,A)$ for all $x\in B$.
Best Answer
Suppose $x$ is in the intersection of these two sets. Then $x \in cl(V)$. Hence the neighborghood $Int(cl(U))$ of $x$ must intersect $V$. But $U \subset V^{c}$ ( $V^{c}$ denoting the complement of $V$) and $V^{c}$ is closed so $\overset {-} U \subset V^{c}$. This implies $Int(cl(U) \subset V^{c}$. we have arrived at a contradiction.