Empty interior of union of cosets? The following question arises from trying to understand Lemma 1.3(ii) of arXiv:math/0405063.  I believe a particular case of the proof (and in fact I think the proof is essentially equivalent to this claim) is:

Let $G$ be a locally compact group.  Let $C,D$ be cosets (not assumed open, closed etc.) each of which has empty interior.  Then $C\cup D$ also has empty interior.

This is not try in general topology, of course: let $C,D$ be the rational, respectively, irrationals, in $\mathbb R$.  However, I cannot decide if being a coset rules out this sort of example.  Is the claim true, and if so, what is a proof?
 A: This is false. Take the (compact abelian) group $G=(\mathbf{Z}/2\mathbf{Z})^\mathbf{N}$ and let $H$ be a dense subgroup of index 2 (there are many, since $G$ has only countably many closed subgroups of index 2 but has $2^c$ subgroups of index $2$, and clearly a subgroup of index 2 is either closed or dense). Then $G=H\cup (G\smallsetminus H)$ and both $H$ and its coset $G\smallsetminus H$ have empty interior.
A: With an additional assumption that there exists an interior point $x\in C\cup D$ such that $x\in C\cap D$, one can prove this fact for any topological group. It doesn't require local compactness.
Note that $e$ is in the interior of $x^{-1}C\cup x^{-1}D$ and both sets are subgroups. Thus, WLOG, $x=e$ and $C$, $D$ are subgroups.
Now pick an open neighbhourhood of $e$ such that $U\subseteq C\cup D$. Pick another open neighbhourhood of $e$ such that $V^2\subseteq U$.
Now an easy algebraic argument shows that $V$ must by a subset of $C$ or $D$. Otherwise, pick $y\in V\setminus D$ and $z\in V\setminus C$. The product $yz$ must be in $U$, so in $C$ or $D$. Both lead to a contradiction.
