 Solution

Choose a finite set $S$ of topological generators of $G$ and let $\Gamma =⟨S⟩$ denote the abstract subgroup it generates. Then $\Gamma$ is dense in $G$. For an open subgroup $H$, the abstract subgroup $K=\Gamma \cap H$ is dense in $H$, since every nonempty open subset of $H$ is open in $G$. Every coset of $H$ meets $\Gamma$, so $[\Gamma :K]=[G:H]=m<\infty$.

Here is the needed Schreier lemma argument explicitly. Choose representatives $t$ in $\Gamma$ for the right cosets $K\backslash \Gamma$, including $1$ for $K$, and write $\overline{\gamma }$ for the representative of $K\gamma$. The finite setis contained in $K$. For $k=s_1\cdots s_r\in K$, set $t_j=\overline{s_1\cdots s_j}$, with $t_0=t_r=1$. Thenand every factor belongs to the displayed finite set. Thus $K$ is finitely generated as an abstract group. Its finite generators topologically generate its closure $H$. Every open subgroup of a finitely generated profinite group is finitely generated.