Following TKe's reference, here is the solution:

Suppose $\mathscr I$ is a set of elements in $G$, a pro-$p$ group.

Denote by $G^*$ the group generated by $G^p$ and $[G,G]$ together. Note that $H^1(G,\mathbb F_p) = \operatorname{Hom}(G,\mathbb F_p) = \operatorname{Hom}(G/G^*,\mathbb Q_p/\mathbb Z_p)$. Then, the claim is that $\mathscr I$ generates $\Gamma$ (topologically) if and only if $\mathscr I$ generates $G/G^*$ as a $\mathbb F_p$ vector space.

This is clearly sufficient to show that $3) \implies 1).$ The crucial idea is the following:

**Lemma:** Any maximal closed subgroup $H \subset G$ is normal of index $p$.

**Proof:** There exists an open subgroup $U$ such that $HU > U$ (otherwise $H=G)$. But then, by Sylow theory, the image of $H$ in $G/U$ is a maximal proper subgroup and is therefore of index $p$ and normal. Moreover, the pullback of this subgroup is in fact exactly $H$ since $H$ is maximal and we are done.

Finally, we have:

**Lemma:** If $G,G'$ are two $pro-p$ groups, then a map $f: G \to G'$ is surjective if and only if $f^*: H^1(G',\mathbb F_p) \to H^1(G,\mathbb F_p)$ is injective.

**Proof:** Clearly, if $f$ is surjective, $f^*$ is injective. Conversely, suppose $f$ is not surjective and let $H = f(G)$. Then, $H$ is closed (by compactness+hausdorff) and hence is contained in a maximal normal subgroup $H'$. $G/H'$ is then isomorphic to $\mathbb F_p$ by the previous lemma and hence gives an element in $H^1(G',\mathbb F_p)$ whose restriction along $f^*$ is $0$ and hence $f^*$ is not injective.

To complete the proof, consider a pro- $p$ group $G$ such that $H^1(G,\mathbb F_p)$ is finite dimensional and let $H$ be a finitely generated subgroup whose image in $G/G^*$ generates $G/G^*$. Then:

$H \to G$ is surjective $\iff$ $H^1(G,\mathbb F_p) \to H^1(H,\mathbb F_p)$ is injective $\iff$ $H/H^* \to G/G^*$ is surjective where the final equivalence is by Poincare duality.

Cohomology of Number Fields, (3.9.1). $\endgroup$ – user19475 Dec 2 '18 at 15:17