Addendum: P. Komjath has pointed out that the following argument is originally due to Erdos and Kunen. See page 136 here.
Claim 1: (Mycielski) Suppose $A$ is a compact subset of plane of positive area. Then there exist perfect sets $P, Q$ such that $Q$ has positive length and $P \times Q \subseteq A$.
Proof: (Martin) Let $M$ be a countable transitive model of ZFC containing a code of $A$. Let $N$ be a forcing extension of $M$ in which continuum is at least $\omega_3$ and Martin's axiom (MA) holds.
Claim 2: Assume MA + $2^{\omega} \geq \omega_3$. For every collection $\{K_{\alpha} : \alpha < \omega_2\}$ of compact sets of positive measure, there is a compact set $ K$ of positive measure such that $ K \subseteq K_{\alpha}$ for at least $ \omega_2$ many $ \alpha$'s.
Proof: WLOG, we may assume that each $K_{\alpha} \subseteq [0, 1]$ with $\mu(K_{\alpha}) > 0.5$. Let $U_{\alpha} = [0, 1] \backslash K_{\alpha}$. Consider the poset $P$ consisting of open subsets of $[0, 1]$ of measure strictly less than $0.5$ ordered by reverse inclusion; i.e., $U \leq V$ iff $V \subseteq U$. $P$ has ccc because the measure algebra is separable. By MA and $2^{\omega} \geq \omega_3$, it follows that every subset of $P$ of size $\omega_2$ has a centered subset of size $\omega_2$. Hence for some $S \in [\omega_2]^{\omega_2}$, any finite union of sets in $\{U_{\alpha}: \alpha \in S\}$ is in $P$. Claim 2 follows.
The claim implies that, in $N$, there is a compact set $Q$ of positive measure such that $W = \{x: Q \subseteq A_x = \{y: (x, y) \in A\}\}$ has size at least $\omega_2$. Since $A$ is closed, $W$ is a coanalytic set hence it is a union of $\omega_1$ many Borel sets. Thus $W$ must contain a prefect set $P$. So $P \times Q \subseteq A$.
Now the statement $(\exists P, Q)(P, Q \text{ are perfect}, Q \text{ has positive measure and }P \times Q \subseteq A)$ is $\Sigma^{1}_2(A)$ so by Shoenfield's absoluteness theorem it holds in $M$ and hence, by $\Pi_1^1$ absoluteness, also in $V$.
Claim 3: Assume $2^{\omega} \geq \omega_2$. Let $X_n$, for $n < \omega$, be pairwise disjoint $\mathbb{Q}$-linearly independent sets of reals. Then $X = \bigcup_n X_n$ has zero inner measure.
Proof: Suppose not and let $K$ be a compact set of positive length contained in $X$. Let $A = \{(x, y) : y \in K + x\}$. By Lemma 1, there are disjoint perfect sets $P, Q$ such that $P \times Q \subseteq A$. Hence for every $x \in P, y \in Q$, $y - x \in K$. Consider the complete bipartite graph $K_{P, Q}$. Let $c$ be an edge coloring of $K_{P, Q}$ such that $c(x, y) = n$ iff $y - x \in X_n$. Let $P' \in [P]^{\omega_1}, Q' \in [Q]^{\omega_2}$. Using Fodor's lemma get $x_0, x_1 \in P'$, $y_0, y_1 \in Q'$ such that $c(x_i, y_j) = n^{\star}$ for every $i, j < 2$. Let $a = y_1 - x_0$, $b = y_1 - x_1$, $c = y_0 - x_1$, $d = y_0 - x_0$. Then, $a, b, c, d \in X_{n^{\star}}, \{a, c\} \cap \{b, d\} = \phi$ but $a-b+c-d = 0$: A contradiction.
