Is it possible to find two disjoint subsets $X$ and $Y$ of $\mathbb{R}$ such that both are dense in $\mathbb{R}$ and both are locally uncountable?
By a locally uncountable set $X \subset \mathbb{R}$, I mean a set which has the property that if I take any nonempty open subset $U$ of $\mathbb{R}$, then $U \cap X$ has cardinality strictly larger than the natural numbers.
Yes. For instance, for each interval with rational endpoints, pick a Cantor set in the interval, and let $X$ be the union of all these Cantor sets. Then $X$ has uncountable intersection with every open interval, but it is meager (or measure zero, if the Cantor sets you chose all have measure zero) so its complement $Y$ has uncountable intersection with every open interval as well.
Here's another construction which gives uncountably many such subsets at once. For each $r\in\mathbb{R}$, let $X_r$ be the set of real numbers whose decimal expansion agrees with $r$'s decimal expansion on all but finitely many of the even digits. Then each $X_r$ has uncountably many points in any open interval, but $X_r$ and $X_s$ are disjoint unless all but finitely many of the even digits of $r$ agree with those of $s$ (in which case $X_r=X_s$).
With heavy use of the axiom of choice, you can get even crazier examples. For instance, you can construct a family of uncountably many disjoint subsets of $\mathbb{R}$, each of which has uncountable intersection with every uncountable closed subset of $\mathbb{R}$. As a sketch of the proof, note that there are $2^{\aleph_0}$ such uncountable closed sets and each has $2^{\aleph_0}$ elements, so you can one by one pick elements to put in each of the sets you're building in a transfinite recursion of length $2^{\aleph_0}$. For more details of this and related constructions, look up "Bernstein sets".