In a Hausdorff topological group, how can I show that every infinite topological group has a countable open subgroup?
I believe the question the poster is trying to ask is, "Why is theorem 3.6 of the article http://link.springer.com/article/10.1023%2FA%3A1010466924961#page-1 true?" Certainly this seems to be something the OP cares about, so I'll address it. (I think this question is borderline for MO, since the answer seems to be rather trivial - feel free to downvote this answer if you think the question is definitely inappropriate, although please say that's why you're downvoting.)
The theorem the OP is asking about is:
The proof of this theorem, in its entirety, is:
The part the OP seems to be asking about is the first sentence. The key is that in the theorem's hypothesis the topological group $(G, \tau_1)$ is assumed to be generated by some $T$-sequence $\alpha$. The reason this matters is that if $(G, \tau_1)$ is determined by $\alpha$, then clearly $A\cup\lbrace 0\rbrace$, where $A$ is the underlying set of $\alpha$, must be open - since $\tau_1$ is maximal among the topologies in which $\alpha$ converges to 0. Now the desired countable open subgroup is just the group generated by $A$.