= Solution
A basis of open sets in the <p-adic integers> is
$$
\boxed{a+p^n\mathbb Z_p\qquad(a\in\mathbb Z_p,\ n\ge0).}
$$
Indeed 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 $u^{-1}\in\mathbb Z_p$. Consequently
$$
p^r\mathbb Z_p=m\mathbb Z_p\subseteq H.
$$
The 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>:
$$
\boxed{H\text{ open}\quad\Longleftrightarrow\quad[\mathbb Z_p:H]<\infty.}
$$
No prior assumption of closedness is required for the finite-index direction.
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