Let $C$ be the set of closed subsets of the reals. The map $\varphi : c \mapsto \mathbb R \setminus c$ is a bijection from $C$ onto the set $O$ of open real subsets of $\mathbb R$. Therefore the cardinality of $C$ is equal to the one of $O$.
Now, it is known that an open subset of the reals is a countable union of disjointed intervals. As the cardinality of the set of real intervals is the continuum, this is also the case of $O$ (the cardinality of $\mathbb R^{\mathbb N}$ is the one of $\mathbb R$). Finally we can conclude that the cardinality of $C$ is equal to the one of $\mathbb R$.