 Solution

Constant sequences give an injection $\mathbb{R}\to {}^{\omega }\mathbb{R}$. For the opposite bound fix an enumeration $(q_m{)}_{m\in \omega }$ of the rational numbers and encode a sequence $(x_n)$ byDistinct sequences differ at some coordinate $n$; density of the rational numbers then makes their encoded sets differ at a pair $(n,m)$. Thus $E$ is injective. The Cartesian product $\omega \times \omega$ is countable; for example the Cantor pairing functionis a bijection with $\omega$. Consequently $\mathcal{P}(\omega \times \omega )$ is equinumerous with $\mathcal{P}(\omega )$, already equinumerous with $\mathbb{R}$. Apply Cantor-Schröder-Bernstein theorem to obtain