I see a proposition which states
The union of countably many open $\sigma$-compact subgroups generates an open $\sigma$-compact subgroup.
The proof in the text is summarized as
The union of countably many open $\sigma$-compact subgroups can be denoted by $(K_{n,j})_{n,j\in\mathbb{N}}$ where $K_{n,j}$ is a compact subset and $\cup_{j\in\mathbb{N}}K_{n,j}$ is an open $\sigma$-compact subgroup. Then the group $G$ generated by the family $(K_{n,j})_{n,j\in\mathbb{N}}$ is therefore $\sigma$-compact.
I guess I don't understand how the $\sigma$-compactness comes. In my opinion, the generated $G$ is just the union of multiplications of finitely many elements of the family $(K_{n,j})_{n,j\in\mathbb{N}}$ and their inverses. I see the multiplication of compact subsets is compact. However,
the family is of size $\aleph_0$, so the number of subsets of the family is $2^{\aleph_0}$ which is uncountable, i.e., the union of those multiplications is a uncountable one, how could we say it is a countable union ?
There are $2^{\aleph_0}$ subsets of the family, but there are only $\aleph_0$ finite subsets of the family. Every element of $G$ is a finite product of points in elements in the family and their inverses, so you only care about finite subsets (or really, finite sequences, since the order may matter).