Without choice, can every homomorphism from a profinite group to a finite group be continuous? In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the axiom of choice: ultrafilters in the first answer, and "every vector space has a basis", in Milne's notes as referenced in the second answer, and used to compute the number of finite-index subgroups in the third answer.
Is it possible to prove the existence of a discontinuous homomorphism from a profinite group to a finite group without the axiom of choice? Instead is it consistent with ZF that there is none?
 A: 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 $H$.
A: There are two common definitions of what it means for a topological group to be profinite, which I’ll distinguish as “formally profinite” and “compact profinite”.

*

*Formally profinite: an inverse limit of finite groups, with the inverse limit topology

*Compact profinite: a compact Hausdorff totally disconnected topological group

By compact I mean that any open cover has a finite subcover. ZF proves that every compact profinite group is formally profinite: it’s a limit of quotients by normal open subgroups. The other implication is not a theorem of ZF. The compactness is equivalent to the Boolean Prime Ideal Theorem.
ZF proves the existence of a discontinuous homomorphism from a certain formally profinite group to a finite group. I don’t know the answer with compact profinite groups, though there are models ruling out some classes of discontinuous map. $\DeclareMathOperator{colim}{colim}$

Let $k=\mathbb F_2.$

Theorem. (ZF) There is a formally profinite group $V$ and a discontinuous group homomorphism $V\to k.$

If we can find a $k$-vector space $V$ such that the natural map $V\to V^{**}$ is not surjective, then we get a homomorphism $V^*\to k$ that is discontinuous when $V^*$ is given the topology of pointwise convergence. This topology makes $V^*$ into a formally profinite group: it’s the inverse limit of duals of finite subspaces of $V.$
The first thing to try is the space of finitely supported $\omega$-sequences, $V_1=\colim_{n\in\omega} k^n.$ The dual space is the direct product $k^\omega.$ If $V_1\to V_1^{**}$ is not surjective we’re done. So we can assume that every linear $k^\omega\to k$ is a dot product with a finitely-supported vector.
The second thing to try is the space of countably supported $\omega_1$-sequences, $V_2=\colim_{\alpha\in\omega_1} k^\alpha.$ Let $e_\alpha\in V_2$ denote the vector with $e_\alpha(\alpha)=1$ and $e_\alpha(\beta)=0$ for $\beta\neq \alpha.$ For each linear $\phi:V_2\to k$ define $I_\phi=\{\alpha:\phi(e_\alpha)\neq 0\}.$
We’ll show that $I_\phi$ is always finite. Fix $\phi$ and let $\beta$ be the supremum of ordinals $\alpha$ such that $\alpha\cap I_\phi$ is finite. By the assumption that every linear $k^{\omega}\to k$ is a dot product with a finitely-supported vector, the restriction of $\phi$ to $k^{\alpha}$ is a dot product with a finitely-supported vector. But its support is just $\beta\cap I_\phi,$ so $I_\phi$ is finite.
Define a linear functional $F:V_2^*\to k$ by taking $\phi$ to $\sum_{\alpha\in I_\phi} \phi(e_\alpha).$ This is not in the image of $V_2\to V_2^{**}$ because $F(x\mapsto x_\alpha)=1$ for all $\alpha.$ In any case we have found a vector space $V$ such that $V\to V^{**}$ is not surjective. If forced to choose we can take $V=V_1\oplus V_2.$ (Old joke)

Compactness can be a sensible notion without the Boolean Prime Ideal Theorem. Tychonoff’s theorem works fine if all its input is well-ordered. In particular, given an inverse system of finite groups $G_i$ where the union of $G_i$ is well-ordered, the inverse limit is compact.
Any non-principal ultrafilter on $I$ defines a discontinuous function on $k^I,$ whether or not $k^I$ is compact. By a result of Blass that I haven’t seen [1], it is consistent with ZF that every ultrafilter is principal. As BS said in their answer, a model of Shelah deals with the second-countable case. So we have two different models ruling out different classes of discontinuous maps. These two results can be combined:

Theorem. It is consistent with ZF that there is no discontinuous map $G\to H$ where $G$ is a product of second-countable compact profinite groups and $H$ is finite. ($G$ itself is not assumed to be compact.)

Pincus and Solovay [2] built on Blass’s argument to construct a model $N$ without ultrafilters and satisfying the handy axiom Dependent Choice. Working in $N$ for now, suppose we have a finite group $H,$ a family of compact second-countable profinite groups $S_i, i\in I$ with product $G=\prod_{i\in I}S_i,$ and a discontinuous homomorphism $\phi:G\to H.$
For each $J\subseteq I$ let $\phi^J:G\to H$ denote the map $\phi^J(g)=\phi(g^J)$ where $g^J_i=g_i$ for $i\in J$ and $g^J_i=1$ for $i\not\in J.$ Let $\mathcal I$ be the ideal of subsets $J\subseteq I$ such that $\phi^J$ is trivial. Let $B$ denote the Boolean algebra $\mathcal P(I)/\mathcal I.$ We’ll consider two cases, and in each case construct a sequence $g_n\in G$ with $\phi(g_n)\neq 1$ such that for every sequence $x\in \{0,1\}^\omega,$ the product $g_0^{x_0}g_1^{x_1}\dots$ converges. (If $G$ is compact, the last condition is equivalent to $g_n\to 1.$)
If $B$ is infinite, pick a countably infinite antichain, and pick disjoint representatives $J_0,J_1,\dots\subset I$ with $\phi^{J_n}$ non-trivial. Also pick $g_n$ with $g_n=g_n^{J_n}$ (i.e. $(g_n)_i=1$ for $i\not\in J_n$) such that $\phi(g_n)\neq 1.$ Any product $g_0^{x_0}g_1^{x_1}\dots$ converges.
Now consider the case that $B$ is finite. For each atom $b$ pick a set $J_b$ representing $b.$ Since $\phi(g)=\phi(\prod_b g^{J_b})=\prod_b \phi^{J_b}(g)$ (with any choice of ordering), some $\phi^{J_b}$ must be discontinuous. Pick one. The ultrafilter $\mathcal P(J_b)\setminus \mathcal I$ is principal because $N$ has no non-principal ultrafilters. So there is an $i\in J_b$ and a homomorphism $\psi:S_i\to H$ with $\phi^{J_b}(g)=\psi(g_i)$ for all $g.$ By second-countability, there is a descending sequence of open normal subgroups $U_0,U_1,\dots$ of $S_i$ such that $\bigcap_j U_j=\{1\}.$ If $\psi$ is trivial on some $U_j$ then $\psi$ is continuous, a contradiction. Otherwise, we can pick $g_n\in U_{k_n}$ with $\psi(g_n)\neq 1$ and $k_n$ strictly increasing to ensure $g_n\to 1.$ Compactness ensures that each limit $g_0^{x_0}g_1^{x_1}\dots$ exists. Embed these $g_n$ in $G$ using the obvious map $S_i\to G.$
We can assume that $H$ is in the ground model $L.$ Define $\psi: 2^\omega\to H$ by $\psi(x)=\phi(g_0^{x_0}g_1^{x_1}\dots).$ Note $\psi(x)\neq \psi(x’)$ whenever $|\{i:x_i\neq x’_i\}|=1.$ The argument that there are no ultrafilters on $\omega$ from [2, Section 1.2] goes through with a small change to use an inequality instead of an equality. (It would be slightly easier to use Cohen forcing here instead of random reals. I've been conservative and avoided changing the model.)
Fix $h\in H.$ Their event $a$ needs to be modified to
$$a_h=\|s(\dot R|v)\in t_h(\dot R|u)\|$$
where $t_h(\dot R|u)$ is a term defining the preimage $\psi^{-1}(\{h\}).$ Their equality $l(a\cap b)=l(b)/2$ needs to be modified to the inequality $l(a_h\cap b)\leq l(b)/2.$ By the measure preserving transformation argument in [2, Section 1.2], all basic opens $b$ satisfy this inequality. By the monotone class theorem, this inequality is also satisfied by $b=a_h,$ giving $l(a_h\cap a_h)\leq l(a_h)/2.$ Hence $l(a_h)=0.$ (Alternatively, apply the Lebesgue density theorem for $2^\omega.$) This holds for all $h,$ contradicting $\bigvee_{h\in H} a_h=1.$

[1] Blass, Andreas, A model without ultrafilters, Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 25, 329-331 (1977). ZBL0365.02054.
[2] Pincus, David; Solovay, Robert M., Definability of measures and ultrafilters, J. Symb. Log. 42, 179-190 (1977). ZBL0384.03030.
