= Solution
There is an <injection> from the <power set> of $\omega$ into the <real numbers> given by
$$
S\longmapsto x_S=\sum_{n=0}^{\infty}\frac{2\mathbf1_S(n)}{3^{n+1}}.
$$
If $S$ and $T$ first differ at $k$, the contribution at $k$ has magnitude $2/3^{k+1}$, while the entire later tail has magnitude at most
$$
\sum_{n>k}\frac2{3^{n+1}}=\frac1{3^{k+1}}.
$$
Their real values are therefore different. This proves <injectivity> directly and avoids any ambiguity of numeral expansions.
Conversely, map a <real number> $x$ to the <set> $\{q\in\mathbb Q:q<x\}$. Distinct <real numbers> give distinct <sets> because the <rational numbers> are dense in the <real line>. Fixing a <bijection> between $\omega$ and the <countable set> $\mathbb Q$ turns this into an <injection> $\mathbb R\to\mathcal P(\omega)$. <Cantor-Schröder-Bernstein theorem> now gives
$$
\boxed{|\mathbb R|=2^{\aleph_0}.}
$$
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