By $\mathbb{R^N}$ I mean the real vector space with the natural componentwise addition and scalar multiplication. Certainly ZFC+(V=L) gives definable bases, but does ZFC?

The answer, if I understand the question correctly, is negative. That is, if we understand "definable" as "can be defined from ordinals [and a real number]", or in simpler words, sets which are in $\sf HOD(\Bbb R)$. Here is a sketch of an argument I suspect is correct. Consider Solovay's model which we get from collapsing an inaccessible cardinal to $\aleph_1$. In that model every set of real numbers definable from an ordinal and a real number has the Baire property. Since I am not going to care about the forcing, $V$ is going to be the model after the collapse. Now suppose that $B$ was a basis and was definable, since $\sf HOD(\Bbb R)$ and $V$ both agree that $B$ must have the cardinality of the real numbers, they also agree that it has more than $2^{\aleph_0}$ permutations. Each of those permutations extends uniquely to an automorphism of $\Bbb{R^N}$. Therefore in $\sf HOD(\Bbb R)$ there are more than $2^{\aleph_0}$ automorphisms of the space. However in $\sf HOD(\Bbb R)$ we have automatic continuity for Polish groups, and therefore every automorphism is continuous. This is a contradiction since there can only be $2^{\aleph_0}$ continuous automorphisms of a Polish space. Therefore $B$ cannot be definable to begin with. One can also get away from the inaccessible cardinal by considering Shelah's model in which not all sets are Lebesgue measurable, but all sets have the Baire property.


