= Solution
For a <profinite group>, its <profinite Frattini subgroup> is
$$
\Phi(G)=\bigcap\{M:M\text{ is a maximal proper open subgroup of }G\}.
$$
Use $G$ for an empty intersection. This is a closed <normal subgroup>, invariant under continuous <group automorphisms>, since these automorphisms permute the maximal open <subgroups>. Let $H=\overline{\langle X\rangle}$. If $H=G$, its image certainly topologically generates $G/\Phi(G)$. Conversely, suppose $H\ne G$. For $g\notin H$, the <open normal subgroup basis of a profinite group> gives $N\trianglelefteq_oG$ with $g\notin HN$: choose a sufficiently small neighbourhood $gN$ disjoint from the closed <subgroup> $H$. The proper <subgroup> $HN/N$ of the <finite group> $G/N$ lies in a maximal proper <subgroup>. Its preimage $M$ is a maximal proper open <subgroup> of $G$, contains $H$, and contains $\Phi(G)$. Since $H\Phi(G)$ is compact and therefore closed, density of the image of $X$ in $G/\Phi(G)$ would force $H\Phi(G)=G$, contradicting $H\Phi(G)\leq M<G$. \b[Generation can therefore be tested modulo the profinite Frattini subgroup.]
For a <finite p-group>, every maximal proper <subgroup> is normal and has index $p$. One proof uses the <normalizer condition for finite p-groups>: maximality forces its normalizer to be the whole <group>, and its simple $p$-group quotient has order $p$. Therefore every such maximal <subgroup> contains powers and <group commutators>. Conversely, the quotient by the algebraic $G^p[G,G]$ is an <elementary abelian p-group>; its nonzero elements are separated by linear maps to $\mathbb F_p$. Their kernels give maximal <subgroups>. This proves the finite <Frattini subgroup> formula.
Now take a <pro-p group> and put $D=\overline{G^p[G,G]}$, retaining the <closure> of the algebraically generated <subgroups>. A maximal open <subgroup> is the preimage of a maximal <subgroup> of a finite $p$-group quotient, so it contains $D$. If $g\notin D$, some finite quotient of the <profinite group> $G/D$ detects $gD$. That quotient is <elementary abelian>, and a linear functional nonzero at $gD$ gives a continuous <group homomorphism> $G\to\mathbb F_p$ whose kernel is a maximal open <subgroup> missing $g$. Hence
$$
\boxed{\Phi(G)=\overline{G^p[G,G]}.}
$$
The <closure> bar is present in the original PDF; the converted TeX fraction is an extraction error.
For the finite-index assertion, first pass from the arbitrary abstract <subgroup> $K$ to its <normal core of a subgroup> $C$. The action on the finitely many <cosets> of $K$ shows $[G:C]<\infty$, without assuming that $K$ or $C$ is closed. Write $m=[G:C]=p^rq$ with $(p,q)=1$, and let $X_m=\{h^m:h\in G\}$. By <Lagrange theorem>, $X_m\subseteq C$. For $N\trianglelefteq_oG$, the quotient $G/N$ is a <finite p-group> of order, say, $p^s$. Choose an integer $a$ with $aq\equiv1\pmod {p^s}$. Then
$$
(g^a)^mN=g^{p^r}N,
$$
so $g^{p^r}\in X_mN$ for every such $N$. The set $X_m$ is the continuous image of the <compact space> $G$, hence is closed. Consequently $g^{p^r}\in\bigcap_NX_mN=X_m\subseteq C$. Every element of the abstract finite quotient $G/C$ thus has order dividing $p^r$. By <Cauchy theorem for groups>, no prime other than $p$ divides its order. Therefore $[G:C]$, and its divisor $[G:K]$, are powers of $p$. \b[This proof does not assume continuity of the finite quotient map.]
To prove openness under finite generation, first establish <commutator collection with fixed generators in nilpotent groups>. If a <nilpotent group> $P$ is generated by $a_1,\ldots,a_d$, every element of $P\prime$ is
$$
[x_1,a_1]\cdots[x_d,a_d].
$$
Induct on the nilpotency class $c$. The assertion is trivial for $c=1$. Apply the induction hypothesis in $P/\gamma_c(P)$; the remaining error lies in the central <subgroup> $\gamma_c(P)=[\gamma_{c-1}(P),P]$. Since commutators in this last layer are central, collecting against words in the $a_i$ expresses the error as $\prod_i[y_i,a_i]$ with $y_i\in\gamma_{c-1}(P)$. Also $[P\prime,\gamma_{c-1}(P)]\leq\gamma_{c+1}(P)=1$. Thus replacing each $x_i$ by $x_iy_i$ multiplies the desired product by precisely that central error, proving the formula.
Let $a_1,\ldots,a_d$ now topologically generate $G$. Every finite continuous quotient is a <nilpotent group>, so the formula holds there with these fixed generators. For $z\in\overline{[G,G]}$, the sets of tuples $(x_1,\ldots,x_d)\in G^d$ solving $z\equiv\prod_i[x_i,a_i]\pmod N$ are nonempty closed subsets of a <compact space>. They have the <finite intersection property>, by replacing finitely many $N$ by their intersection. A common tuple solves the equality in $G$. It follows that the algebraic <commutator subgroup> is the compact image of $G^d$ under this product map, and is closed.
Its quotient $A=G/[G,G]$ is a topologically finitely generated abelian <pro-p group>. The continuous map $\mathbb Z_p^d\to A$ taking $(\lambda_i)$ to $\prod_i\bar a_i^{\lambda_i}$ is onto, because its compact image contains a dense generating <subgroup>. Therefore $A/pA$ is generated as an <elementary abelian group> by at most $d$ elements, and $pA$ is compact. The preimage $G^p[G,G]$ of $pA$ is closed of finite index, hence open: its complement is a finite union of closed <cosets>. The finite-index result and the maximal-subgroup property of <finite p-groups> show that any abstract index-$p$ <subgroup> of $G$ is normal. It contains this open <subgroup>, so is itself open.
Finally, a finite-index normal <subgroup> $C$ has finite $p$-group quotient. If $[G:C]>1$, choose a normal index-$p$ <subgroup> in that quotient and call its preimage $H$. The preceding argument makes $H$ open. An open <subgroup> of a finitely generated <pro-p group> is finitely generated: intersect a finitely generated dense abstract <subgroup> with $H$, use the <Schreier lemma>, and observe that the intersection is dense in the open <subgroup> $H$. Induction on $[G:C]$ now makes $C$ open in $H$, hence in $G$. Returning to the <normal core of a subgroup>, $K$ is a union of open <cosets> of $C$. \b[Every finite-index subgroup of a finitely generated pro-p group is open.]
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