= 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> $(x_n)$ by
$$
E((x_n))=\{(n,m)\in\omega\times\omega:q_m<x_n\}.
$$
Distinct <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 function>
$$
\pi(n,m)=\frac{(n+m)(n+m+1)}2+m
$$
is 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
$$
\boxed{|{}^{\omega}\mathbb R|=2^{\aleph_0}.}
$$
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