Non-Borel subspace of Banach space Let $X$ be a separable Banach space, $M \subset X$ a linear subspace.  Must $M$ be a Borel set in $X$?
I believe the answer is "no," since I have seen authors who are careful to talk about "Borel subspaces".  But I have not been able to find a counterexample.
If the answer is indeed "no", does every infinite-dimensional separable Banach space contain a non-Borel dense linear subspace?
 A: Similar question was discussed before and answered (in particular to Bill's question in the comments) 
Are proper linear subspaces of Banach spaces always meager?
A: This is an answer, but not the "right" answer".  
Presumably you mean that $X$ is infinite dimensional and hence has Hamel dimension the continuum $c$.  For every subset of a given Hamel basis you get the linear subspace spanned by the subset, and these subspaces are different for different subsets of the basis.  Thus $X$ has $2^c$ linear subspaces but only $c$ Borel sets (since $X$ is separable).
EDIT: I just noticed the second question.  Consider subsets of the basis that all contain one fixed countable subset whose span is dense.
A: Let $X$ be a separable Banach space.  Any linear subspace with the property of Baire (in particular, any linear subspace that is a Borel set) is closed.  According to the Axiom of Choice, if $X$ is also infinite-dimensional, then there are discontinuous linear functionals, and their kernels are non-closed (and hence non-Borel) linear subspaces---of course these are dense.
EDIT
The above is clearly wrong, as the comments show.  I could try to salvage something maybe saying $G_\delta$-set instead of property of Baire, but that would not answer the original question here.
