The axiom of choice is, for ZF, equivalent to the statement that every vector space has a basis. The implication of AoC by the existence of a basis for any vector space is shown in this paper.
The proof of that theorem can be summed up as following: One adjoins all members of the sets $X_i$ from which one needs choice as variables to an arbitrary field $k$ creating the field of rational functions $k(X)$ over that field. Then a subfield $K$ of that field is defined through a variation of homogeneity that considers the index of the $X_i$ from which variables are occurring and finally the span of the variables as a vector space $V$ over $K$ is considered. A basis for $V$ over $K$ is chosen, and the construction is such that for all $x \in X_i$, the same set of basis elements occurs with non-zero coefficient in their $K$-representation. This fact is then used together with a similarity of the coefficients occuring to pick a unique $x_i \in X_i$ and an element was chosen as desired.
Now, this comment made me curious if there can be a $\mathbb{Q}$-basis of $\mathbb{R}$ when choice fails at size continuum or below.
That separates into three potentially hard (or, perhaps more likely, easily negatively answered) questions:
a) Can we conclude anything from the existence of one specific basis? The construction of the critical field in the proof relies on the structure of the set family we choose from, and so it seems that these fields will not be isomorphic for different families of sets. But that doesn't say there cannot be a different proof where they are. Or it might also be possible to choose that one guaranteed basis for a family such we can imply choice for all other set families of the same cardinality - but then, does such a family exist? Any direct proof of the existence of well-orderings I know use multiple instances of choice, but perhaps one carefully chosen one can be enough?
b) Assuming a) is positively answered, can we make any conclusions about lower cardinalities? Or, under ZF more strongly, can we actually say something about all subsets even if they are not comparable - can they still exist if we have choice for all families of sets of cardinality $\frak c$?
c) Even if a) is answered positively, could a $\mathbb{Q}$-basis for $\mathbb{R}$ be used as this basis? This would then obviously require a different proof.
To make the question precise, consider three versions of "hold up to $\frak c$":
Do we have choice for families of sets where both the index set and all sets are subsets of $\mathbb{R}$?
And, not necessarily equivalently, do we have choice for families of sets where both the index set and all sets have defined cardinalities $\le\frak c$ or perhaps $<\frak c$?