These notes written by Julien Melleray help us to solve the problem. I just state the results which will help in our case.
Lemma 3.3 (Pettis) Let $G$ a Polish group. For $A\subset G$, define $U(A)$ as the biggest open set $V$ such that $A$ is comeagre $V$. For any subsets $A$ and $B$ of $G$, we have $$U(A)\cdot U(B)\subset A\cdot B.$$
As a consequence:
Theorem 3.4 Let $G$ a Polish group, and $A$ a Baire measurable non-meagre subset of $G$. Then $e$, the neutral element, belongs to the interior of $A\cdot A^{-1}$.
Back to the problem. Of course, $\Bbb R$ with the addition is a Polish group, and Borel/Baire $\sigma$-algebras are the same. Let $H$ a subgroup of $\Bbb R$ which is non-meagre and Borel measurable. It's Baire measurable. By the last theorem, $e$ belongs to the interior of $H\cdot H^{-1}=H$ as $H$ is a sub-group.
It's well-know that the subgroups of $\Bbb R$ are either of the form $a\Bbb Z$ (hence meagre) or dense. So we have a subgroup $H$ which is dense and has non-empty interior, say $(-r,r)$. Let $x\in \Bbb R$, and $x'\in H$ such that $|x-x'|\lt r$. Then $x-x'\in H$ and $x\in H$.
