 Solution

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 $e$. DefineThe identity belongs to $S$, so these powers are increasing. Symmetry gives $H^{-1}=H$, and $S^m{S}^n=S^{m+n}$ 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.