Here are three properties a topological group $G$ might have.
(1) The group is topologically finitely generated.
(2) All abstract subgroups with finite index are open.
(3) There are finitely many abstract subgroups with each index.
In 1975, Serre showed that (1) implies (2) for pro-$p$ groups (a proof is in Sect. 4.3.4 of
J. S. Wilson, Profinite Groups, Oxford Univ. Press, Oxford, 1998) and he conjectured that (1) implies (2) for all profinite groups. This conjecture was proved in 2007 by N. Nikolov and D. Segal in 2007 using the classification of finite simple groups. (N. Nikolov and D. Segal, On finitely generated profinite groups. I. Strong completeness and uniform bounds, Ann. of Math. 165 (2007), 171--238 and On finitely generated profinite groups. II. Products in quasisimple groups, Ann. of Math. 165 (2007), 239--273.
H. L. Peterson (Discontinuous characters and subgroups of finite index, Pacific J. Math. 44 (1973), 683--691) showed (2) implies (3) for compact Hausdorff groups and M. P.
Anderson (Subgroups of finite index in profinite groups, Pacific J. Math. 62 (1976),
19--28) showed (3) implies (2) for pro-solvable groups. Therefore (2) and (3) are equivalent for pro-solvable groups, thus in particular
for pro-$p$ groups. That (3) implies (2) for all compact Hausdorff groups
(thus all profinite groups) is due to J. S. Wilson (Lemma 6 in Groups satisfying the maximal condition for normal subgroups, Math. Z. 118 (1970), 107--114), while
M. G. Smith and J. S. Wilson (On subgroups of finite index in compact Hausdorff groups,
Arch. Math. 80 (2003), 123--129) reproved the equivalence of (2) and (3) for
compact Hausdorff groups.
At this point we see that (1) implies (2) for profinite groups and (2) and (3) are equivalent for compact Hausdorff groups. To complete the circuit,
it is natural to guess that (3) implies (1) for profinite groups (and perhaps even all
compact Hausdorff groups).
We will check (3) implies (1) for pro-$p$ groups, so (1), (2), and (3) are equivalent when
$G$ is a pro-$p$ group.
Let $G$ be a pro-$p$ group satisfying (3). A maximal open subgroup of a pro-$p$
group has index $p$, so (3) implies there are finitely many maximal open subgroups of $G$. The intersection of finitely many open subgroups is open, so the intersection of the maximal open subgroups of $G$ is open. This intersection is the Frattini subgroup $\Phi(G)$, so $\Phi(G)$ is open in $G$. It follows from Prop. 2.8.10 in
L. Ribes and P. Zalesskii, Profinite Groups, Springer--Verlag, Berlin, 2000 that
$G$ is topologically finitely generated.
Example taken from Ribes and Zalesskii's book "Profinite groups". Take an infinite set $I$ and a finite group $T$. You can let $G$ be the profinite group $\prod_I T$. Denote its elements by $(g_i)_{i\in I}$. Let $\mathcal F$ be an ultrafilter which contains the filter of all cofinite subsets of $I$. If you denote $H$ to be the subgroup of elements with $\lbrace i\in I \mid g_i=1\rbrace\in \mathcal F$, it is clear that $H$ is proper normal and that it is not open because it is dense and has finite index $|T|$.
To show that $H$ has index $|T|$ in $G$ consider all elements $a_t=(t,t,\dots)$ for $t\in T$. For any $g\in G$, consider $I_t=\lbrace i\in I \mid g_i=t\rbrace$. Since we have $\bigcup_{t\in T} I_t=I$
then $I_t\in \mathcal F$ for some $t$, and therefore $ga_t^{-1}\in H$.
Best Answer
Saharon Shelah (elaborating on Solovay) proved that it is equiconsistent with ZF that there exists a model of ZF+DC (DC=dependent choice, ) in which all sets of reals have the Baire property. And indeed all subsets of any Polish space have this property.
Now if $G$ is a second countable (=metrizable) profinite group, it is a Polish space (indeed a Cantor), and if $H$ is a non-open finite index subgroup, $H$ cannot have the Baire property, by a standard argument (it would have to be meager, which is absurd).
Hence it is consistent with ZF(+DC) that there is no discontinuous homomorphism from a second countable profinite group to a finite group.
On the other hand, I don't know if one can reduce the general case to the second countable one, e.g. by finding a countable intersection of open normal subgroups inside $H$.