This is not really a new solution, but just a way to see one can come up with Jim's answer. The problem is equivalent to finding an equivalence relation on $\mathbb{R}$ such that each equivalence class is dense and there are $2^{\aleph_0}$ equivalence classes. To see this, suppose you have such an equivalence class $\equiv$. Consider the natural map $\pi: \mathbb{R} \to \mathbb{R}/\equiv$. Clearly the pre-image of every point is dense and now, you can post-compose $\pi$ with a bijection between $\mathbb{R}/\equiv $ and $\mathbb{R}$. The converse is similar.

Now, as $\mathbb{R}$ has the structure of an additive group, one can define equivalence relations by using subgroups: $x \equiv y$ iff $x-y$ is in a given subgroup $H$. Now if you use $H=\mathbb{Q}$, then you get the answer given by Jim Belk. You can use $H$ to be the subgroup of $2$-adic rationals. Then the $x$ is equivalent to $y$, if all but finitely many of their binary digits are equal, from which you can define