Let $G$ be a locally compact group. The union of countably many open $\sigma$-compact subgroups of $G$ generates an open $\sigma$-compact subgroup.
For each $n \in \Bbb{N}$, let $L_n = \bigcup_{j=1} K_{n,j}$ be an open subgroup, where $K_{n,j}$ is compact. I was able to prove that $\langle \bigcup_{n=1}^\infty L_n \rangle = \langle \bigcup_{n=1}^\infty \bigcup_{j=1}^\infty K_{n,j} \rangle$ is open; and I'm pretty certain it's true that
$$\langle \bigcup_{n=1}^\infty L_n \rangle = \bigcup_{t=1}^\infty \bigcup_{(n_1,...,n_t),(j_1,...,j_t) \in \Bbb{N}^k} K_{n_1,j_1} ... K_{n_t,j_t}$$
It's clear that $K_{n_1,j_1} ... K_{n_t,j_t}$ is compact. However, it isn't clear that that nasty double union is a single countable union of compact sets. How does one argue for that conclusion?