The continuum is equinumerous with the power set of the naturals

The inclusion $C\hookrightarrow \mathbb{R}$ composed with [L1] gives an injection ${}^{\omega }2\hookrightarrow \mathbb{R}$. This construction uses no choice.

For each real $x$, put $D_x=\{q\in \mathbb{Q}:q<x\}$. If $x<y$, choose a rational $q$ with $x<q<y$ by [L2]. Then $q\in D_y\setminus D_x$, so $x\mapsto D_x$ injects $\mathbb{R}$ into $\mathcal{P}(\mathbb{Q})$. A fixed bijection $\mathbb{Q}\approx \mathbb{N}$ and [L3] give an injection $\mathbb{R}\hookrightarrow \mathcal{P}(\mathbb{N})$. No family of choices is made: only the existence of one separating rational is used to prove injectivity.

Sending $S\subseteq \mathbb{N}$ to its characteristic function ${\mathbf{1}}_S:\omega \to 2$ is a bijection $\mathcal{P}(\mathbb{N})\approx {}^{\omega }2$, with inverse $b\mapsto b^{-1}(\{1\})$. Combine it with steps 1.1 and 1.2. There are injections in both directions between $\mathbb{R}$ and ${}^{\omega }2$, so Schröder–Bernstein [L3] gives $\mathbb{R}\approx {}^{\omega }2\approx \mathcal{P}(\mathbb{N})$ in ZF.

Now assume Choice. By [L4], the equinumerous sets in step 2.1 have equal cardinalities, while $\mathrm{\mid }\mathbb{N}\mathrm{\mid }={\mathrm{\aleph }}_0$ and $\mathrm{\mid }\mathcal{P}(\mathbb{N})\mathrm{\mid }=2^{\mathrm{\mid }\mathbb{N}\mathrm{\mid }}$. Hence $\mathrm{\mid }\mathbb{R}\mathrm{\mid }=2^{{\mathrm{\aleph }}_0}=\mathrm{\mid }\mathcal{P}(\mathbb{N})\mathrm{\mid }$.