Every locally compact Hausdorff group has an open sigma-compact subgroup

Choose a compact neighbourhood $K$ of $e$, which exists by [F1]. Its inverse $K^{-1}$ is compact by [F2] and [F3], so $K\times K^{-1}$ is compact; the continuous multiplication map sends it onto $U=K{K}^{-1}$, hence $U$ is compact by [F2] and [F3]. Since $e\in K$, we have $K\subseteq U$, so $U$ contains an open neighbourhood of $e$ by [F1]. Finally, [F4] gives $(x{y}^{-1}{)}^{-1}=y{x}^{-1}$, and relabeling $x,y\in K$ shows $U^{-1}=U$. Thus $U$ is a symmetric compact neighbourhood of $e$.

We have $e\in U^0$, $U=U^{-1}$, and $U^m{U}^n=U^{m+n}$ for all $m,n\in \mathbb{N}$, so $H={\bigcup }_n{U}^n$ contains $e$, is closed under products, and is closed under inverses; hence it is a subgroup. Each $U^n$ is compact: $U^0=\{e\}$ is compact, and if $U^n$ is compact, then $U^{n+1}$ is the continuous image of the compact product $U^n\times U$ under multiplication, so it is compact by [F3]. Induction [F5] proves this for every $n$, and the displayed $\mathbb{N}$-indexed union makes $H$ sigma-compact.

By [F1], choose an open neighbourhood $V$ of $e$ contained in $K$; then $V\subseteq U\subseteq H$. For every $h\in H$, [F6] makes $hV$ open, and the subgroup property gives $hV\subseteq H$. Since $e\in V$, each $h\in H$ lies in $hV$, so $H={\bigcup }_{h\in H}hV$ is open. The existence of $K$ follows from local compactness, completing the claim for every locally compact Hausdorff group.