Here is a positive answer, unfortunately it is predicated on a hypothesis which is widely believed to be false: that the existence of inaccessible cardinals is inconsistent with ZFC. Stated in a more digestible manner, the argument below shows that some large cardinal hypotheses are necessary to obtain a negative answer to the question.

Let $V$ be a model of $ZFC$, let $\mathfrak{c}^V$ denote the cardinality of the continuum in $V$ and let $\aleph_1^V$ denote the first uncountable ordinal in $V$. In $L$, $\aleph_1^V$ must be an uncountable regular cardinal and, since there aren't any inaccessible cardinals, it must be a successor cardinal. Let $\kappa$ denote the cardinal in $L$ such that $(\kappa^+)^L = \aleph_1^V$; note that $\kappa$ is definable in $V$. In $V$, $\kappa$ is countable, so there are reals $r$ in $V$ such that $\kappa$ is countable in $L[r]$ and for such reals we necessarily have $\aleph_1^{L[r]} = \aleph_1^V$. Let $K$ be the set of all reals such that $\aleph_1^{L[r]} = \aleph_1^V$; note that $K$ is definable in $V$ and $K$ has size $\mathfrak{c}^V$. For each $r \in K$, $X_r = \mathbb{R}^{L[r]}$ is a set of reals with a canonical wellordering of order type $\aleph_1^{L[r]} = \aleph_1^{V}$.

Following Raisonnier [*A mathematical proof of S. Shelah's theorem on the measure problem and related results*, Israel J. Math. 48 (1984), no. 1, 48–56; MR0768265; DOI:10.1007/BF02760523], we can associate to each $r \in K$ a filter $F_r$ and a collection of sets $\mathcal{H}_r$ of size at most $\mathfrak{c}^V$ such that either $F_r$ is rapid and hence not measurable, or at least one element of $\mathcal{H}_r$ is not measurable. (The set $\mathcal{H}_r$ consists of all sets that Raisonnier denotes $\tilde{H}(X_r)$. These are associated to $G_\delta$ sets $H \subseteq \mathbb{R}\times\mathbb{R}$ with null sections; since there are continuum many such $G_\delta$ sets, it follows that $|\mathcal{H}_r| \leq \mathfrak{c}^V$.)

Putting all this together, under the unlikely assumption that inaccessible cardinals provably do not exist in ZFC, in any model $V$ of ZFC, the definable set $$\{F_r : r \in K\} \cup \bigcup_{r \in K} \mathcal{H}_r$$ has size $\mathfrak{c}^V$ and it must contain a non measurable set.

notLebesgue measurable"? $\endgroup$ – Trevor Wilson May 3 '14 at 6:38definable. There are models of $\mathsf{ZFC}$ where every definable set of reals is Lebesgue measurable. $\endgroup$ – Trevor Wilson May 4 '14 at 0:0319more comments