No, it is not possible. It is consistent with ZF without choice that
the reals are the countable union of countable sets. (*)
From this it follows that all sets of reals are Borel. Of course, the "axiom" (*) makes it impossible to do any analysis. As soon as one allows the bit of choice that it is typically used to set up classical analysis as one is used to (mostly countable choice, but DC seems needed for Radon-Nikodym), one can implement the arguments needed to show
(**) The usual hierarchy of Borel sets (obtained by first taking open sets, then complements, then countable unions of these, then complements, etc) does not terminate before stage $\omega_1$ (this is a kind of diagonal argument).
Logicians call the sets obtained this way $\Delta^1_1$. They are in general a subcollection of the Borel sets. To show that they are all the Borel sets requires a bit of choice (One needs that $\omega_1$ is regular).
There is actually a nice result of Suslin relevant here. He proved that the Borel sets are precisely the $\Delta^1_1$ sets: These are the sets that are simultaneously the continuous image of a Borel set ($\Sigma^1_1$ sets), and the complement of such a set ($\Pi^1_1$ sets).
That there are $\Pi^1_1$ sets that are not $\Delta^1_1$ (and therefore, via a bit of choice, not Borel) is again a result of Suslin. He also showed that any $\Sigma^1_1$ set is either countable, or contains a copy of Cantor's set and therefore has the same size as the reals. His example of a $\Sigma^1_1$ not $\Delta^1_1$ set uses logic (a bit of effective descriptive set theory), and nowadays is more common to use the example of the $\Pi^1_1$ set WO mentioned by Joel, which is not $\Delta^1_1$ by what logicians call a boundedness argument.
A nice reference for some of these issues is the book Mansfield-Weitkamp, Recursive Aspects of Descriptive Set Theory, Oxford University Press, Oxford (1985).
H. Läuchli [Auswahlaxiom in der Algebra, Comment. Math. Helv. 37 1962/1963, MR143705] constructed a permutation model wherein there is a vector space which is not finite dimensional and such that all of its proper subspaces are finite dimensional. This is an example of a vector space with the property you want.
Best Answer
This earlier answer of mine shows how to get an uncountable $\mathbb{Q}$-independent subset of $\mathbb{R}$ in ZF. This set is not a Hamel basis so the $\mathbb{Q}$-span of this set is as required.