I am able to see why the closed nowhere dense subsets of $\mathbb R$ are equinumerous with $\mathbb R$: Every closed nowhere dense subset of $\mathbb R$ is the boundary of an open set (namely its compliment). $\mathbb R$ has $2^{\aleph_0}$ many open sets, so there are at most $2^{\aleph_0}$ many closed nowhere dense subsets of $\mathbb R$. Every Cantor set is a closed nowhere dense subset of $\mathbb R$, and there are $2^{\aleph_0}$ many different Cantor sets, so there are precisely $2^{\aleph_0}$ many closed nowhere dense subsets of $\mathbb R$.
My next thought was to take closures, but this is not injective. What am I missing?
Just a hint, please!
As pointed out in the comments, there are more than continuum many nowhere dense subsets of $\mathbb R$. In particular, the Cantor set $C$ has the same size as the reals, and any subset of $C$ is nowhere dense, so there are at least $|2^{C}| = 2^{2^{\aleph_0}} > 2^{\aleph_0}$ such sets.