Every locally compact topological group is normal space

 

Let $U$ be a neighbourhood of the unity element $e$ of the locally compact topological group $G$. $U$ can be taken open, relatively compact and symmetric, since $U=V \cap V^{-1}$ for some open neighbouhood $V$ of $e$. Let $H=\cup_{n\in N} U^{n} = \cup_{n\in N} \bar{U}^{n}$ be an open, therefore closed subgroup of $G$. [: It is easy to see $\cup_{n\in N} U^{n}$ is an open subgroup. Moreover, $\cup_{n\in N} U^{n}= \overline{\cup_{n\in N} U^{n}}\supset \cup_{n\in N} \overline{U^{n}}\supset \cup_{n\in N} U^{n}$]. It is a countable union of the compact sets $\bar{U}^{n}$, so it is a σ-compact space, therefore Lindelof, therefore normal, since Lindelof and regularity gives normality. And topological groups are always regular: Let $U$ be an open neighbourhood of $e$ and $V$ a symmetric open  neighbourhood of $e$ such that $V^{2}\subset U$. Let $x\in \bar{V}$. Then $xV\cap V \neq \emptyset$, so there exist $v_{1},v_{2} \in V$ with $xv_{1}=v_{2}$, which gives $x=v_{2}(v_{1})^{-1}\in VV^{-1}=VV\subset U$. So, $x\in \bar{V}\subset U$ showing the regularity. 

If $H=G$, the proof is finished. Otherwise, let $a\in G\diagdown H$. Then $e\notin aH$: if $e=ah$ for some $h\in H$, then $a=h^{-1}\in H$. Thus, any two of the sets $aH$, $a\in G$ are disjoint or identical, namely $aH\cap bH=\emptyset$ or $aH=bH$, $a,b\in G$. So, $\{aH: a\in G\}$ is a cover of $G$ by disjoint, open-closed normal subspaces. 

 Let $A,B$ be disjoint closed subsets of$G$. Let $S=\{a\in G: aH\cap A\neq \emptyset\}$, $T=\{a\in G: aH\cap B\neq \emptyset\}$. If $a\in S\cap T$ then $(aH\cap A)\cap (aH\cap B) =  \emptyset$ and $aH\cap A, aH\cap B$ closed in $aH$, since $aH$ is closed too.  So, in the normal space $aH$ there are disjoint open sets $V_{a}, U_{a}$ with $aH\cap A \subset V_{a}, aH\cap B \subset U_{a}$. Since $aH$ is open in $G$, $V_{a}, U_{a}$ are also open in $G$. Furthermore, in case $aH, a'H$ are disjoint for $a\neq a' \in G$ then obviously $V_{a}, U_{a'}$ are disjoint too. Thus, $\cup_{a\in S} V_{a}$ and $\cup_{a\in T} U_{a}$ are the open disjoint sets containing $A, B$.