Others have commented that being dense with a dense complement is not enough. In fact, totally imperfect sets of reals and their complements satisfy a much stronger type of density.
THEOREM: Let $X$ be a totally imperfect subset of the reals $\mathbb{R}$. Then, for each open interval $(a,b)$, we have:
(a) The outer Lebesgue measure of $X \cap (a,b)$ is $b - a$.
(b) The outer Lebesgue measure of $(\mathbb{R}\setminus X) \cap (a,b)$ is $b - a$.
PROOF: If the inner Lebesgue measure of $X \cap (a,b)$ were positive, then this intersection would contain a Lebesgue measurable set of positive measure, hence a closed set of positive Lebesgue measure, hence a perfect set of positive Lebesgue measure (Cantor-Bendixson), which contradicts $\mathbb{R}\setminus X$ having nonempty intersection with every perfect set. Therefore, the inner Lebesgue measure of $X \cap (a,b)$ is zero. The same argument works for $\mathbb{R}\setminus X$.
REMARK: This proves that every totally imperfect set in the reals fails to be Lebesgue measurable, even nowhere locally Lebesgue measurable, and in a rather strong way since finite additivity is enough to get a contradiction.
(FYI, I haven't had a chance to read about how to post math symbols like 'intersect' yet, and copying/pasting from other posts didn't work. Others should feel free to edit my comments using the appropriate math symbols.)