Is there a group with countably many subgroups, but is not countable in ZF? Inspired by this question, although I don't think it was the OP's intention, hence this separate question: Is there a group $G$ with countably many subgroups, but is a not a countable group itself in $\mathrm{ZF}$?
In $\mathrm{ZFC}$ we can look at the cyclic subgroups of $G$ and "estimate" the number of elements in the group, to conclude that $G$ is countable. But this ends up not going through in $\mathrm{ZF}$ since a countable union of finite sets does not have to be countable, in particular it is known that a countable union of two-element sets does not have to be countable. 
So a possible way to construct such a uncountable group (although I am not saying this is a good way to go, I have no idea) is start with a collection $\{ A_i \mid i \in \mathbb{N} \}$ where $A_i$ are pairs, whose union is not a countable set and note that every torsion-free cyclic group has two natural generators, so conceivably there could be a torsion-free group with $A_i$ the natural generating set for a cyclic groups ("$1,-1$" but we could not actually define such a function all $A_i$ without the axiom of choice). Then the constructions would have to make sure there are only countable many subgroups (this seems difficult and would take a lot of care)
An interesting paper "On the number of Russell's socks or $2+2+2+\cdots = ?$" 
by Horst Herrlich, Eleftherios Tachtsis discusses some of the ideas around countable unions of pairs.
 A: I believe the answer is yes, as follows:
Start with a model of ZF+atoms, $M$, with a set of atoms $A$ which forms a group isomorphic to $\mathbb{D}/\mathbb{Z}$, where $\mathbb{D}$ is the set of dyadic fractions: $\mathbb{D}=\{{p\over 2^k}: p, k\in\mathbb{Z}\}$. Let $G$ be the group of automorphisms of $A$, and consider the symmetric submodel $N$ of $M$ corresponding to the filter of finite supports on $G$. Then in $N$, $A$ is no longer countable, since there are nontrivial automorphisms of $\mathbb{D}/\mathbb{Z}$ fixing arbitrary finite sets; but the only subgroups of $\mathbb{D}/\mathbb{Z}$, other than the whole thing, are those of the form $$\{x: 2^kx=0\}$$ for some fixed $k\in\mathbb{N}$. This provides an explicit bijection - in the original universe, $M$ - between the subgroups of $A$ and $\omega$. Now, passing to $N$, we get no additional subgroups of $A$, and the map described above is symmetric; so in $N$, $A$ has only countably many subgroups.
Meanwhile, the statements "$A$ is uncountable" and "$A$ has countably many finitely generated subgroups" are each "bounded," so we may apply the Jech-Sochor theorem to push this construction into the ZF-setting.
