 Solution

A basis of open sets in the p-adic integers isIndeed this is the set of $x$ with $v_p(x-a)\ge n$, equivalently $|x-a{|}_p<p^{-(n-1)}$; these neighborhoods become arbitrarily small. Reduction identifies ${\mathbb{Z}}_p/p^n{\mathbb{Z}}_p$ with $\mathbb{Z}/p^n\mathbb{Z}$, so each has index $p^n$.

If a subgroup $H\le ({\mathbb{Z}}_p,+)$ is open, a neighborhood of zero of the displayed form lies inside $H$. Thus $p^n{\mathbb{Z}}_p\subseteq H$ for some $n$, and $[{\mathbb{Z}}_p:H]\le p^n$ is finite.

Conversely suppose the abstract index is the finite integer $m$. The quotient is a finite abelian group of order $m$, so Lagrange's theorem gives $mx\in H$ for every $x\in {\mathbb{Z}}_p$. Write $m=p^{r}u$ with $p\nmid u$. Multiplication by the p-adic unit $u$ is a bijection of ${\mathbb{Z}}_p$, its inverse being multiplication by $u^{-1}\in {\mathbb{Z}}_p$. ConsequentlyThe subgroup $H$ is a union of open cosets of $p^r{\mathbb{Z}}_p$, and is therefore open. This proves the characterization of open subgroups of the p-adic integers:No prior assumption of closedness is required for the finite-index direction.