= Solution
\b[True.] Choose a compact neighbourhood $K$ of the identity in the <locally compact group> and put $S=K\cup K^{-1}\cup\{e\}$. Inversion is a <homeomorphism>, so $S$ is compact; it is symmetric and still a neighbourhood of $e$. Define
$$
\boxed{H=\bigcup_{n=1}^{\infty}S^n.}
$$
The identity belongs to $S$, so these powers are increasing. Symmetry gives $H^{-1}=H$, and $S^mS^n=S^{m+n}$ gives closure under multiplication; hence $H$ is a <subgroup>. Each $S^n$ is compact, being the continuous image of the finite product of compact spaces. Thus $H$ is a <sigma-compact space>. Finally $S$ contains some open identity neighbourhood $U$, and $H=\bigcup_{h\in H}hU$ is open. This constructs the required <sigma-compact open subgroup of a locally compact group>. The argument also works without Hausdorffness if local compactness means possession of compact neighbourhoods.
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