The existence of such subgroups implies the existence of a non-measurable set; simply intersect each of the cosets with $[0,1]$. The results will all have equal outer measure, but their union will be all of $[0,1]$, and you only have countably many of them. As a result, they are non-measurable.
So, for example, in the Solovay model, these subgroups do not exist. However, with choice, you can let $B$ be a Hamel basis, then using that to generate a surjective homomorphism $F$ from $\mathbb{R}$ to $\mathbb{Q}$ (as in here) and consider all of the elements $x$ such that $F(x)=0$. This forms a subgroup under addition (since its a homomorphism) and then its cosets are those that contain all $y$ such that $F(y)=q$ for different rationals $q$; as a result, the group has countably infinite index.
Now, as shown in the question linked to above, the existence of such a subgroup in $\text{ZF+DC}$ is undecidable (models where all sets of reals are Borel for example). However, the existence of a Hamel basis is enough to prove the existence of one of these subgroups.
I believe that the existence of one of these subgroups is not equivalent to the existence of a Hamel basis under $\text{ZF}$. Perhaps it is possible for there to be a model in which a countably-infinite subgroup exists but its cosets do not correspond to different rational numbers as above, in which case theoretically the Hamel basis would be unnecessary.
Question: How much choice is required for there to be a subgroup of the real additive group with countably-infinite index?
