$\mathbb{R} \approx \mathcal{P}(\mathbb{N})$ in ZF, by the Cantor set for one injection and by the cuts $\{q \in \mathbb{Q} : q < x\}$ for the other; so $\lvert \mathbb{R} \rvert = 2^{\aleph_0}$ under the Axiom of Choice

Write ${}^{\omega}2$ for the set of functions $\omega \to 2 = \{0,1\}$, the $2$ here being the von Neumann natural number, and $\mathcal{P}(\mathbb{N})$ for the power set of $\mathbb{N} = \omega$ (The natural numbers $\mathbb{N}$ (von Neumann)). Then:

(a) In ZF, with no choice principle:

$\mathbb{R} \;\approx\; {}^{\omega}2 \;\approx\; \mathcal{P}(\mathbb{N})$

(Equinumerous sets, $A \approx B$ and $A \preceq B$, The real numbers).

(b) Assuming the Axiom of Choice (The Axiom of Choice):

$\lvert \mathbb{R} \rvert \;=\; 2^{\aleph_0} \;=\; \lvert \mathcal{P}(\mathbb{N}) \rvert$

(Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation $\kappa^{\lambda}$, and why they are written apart from the ordinal operations, The successor cardinal $\kappa^{+}$, the alephs $\aleph_\alpha$, the beths $\beth_\alpha$, successor and limit cardinals, and the identifications $\aleph_0 = \omega$ and $\aleph_1 = \omega_1$).

The two injections are the classical ones and neither uses a binary expansion. One direction is the Cantor set: The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$, and this gives a bijection with $\{0,1\}^{\mathbb{N}}$ already supplies a bijection from the sequences with values in $\{0,1\}$ onto the Cantor set $C \subseteq \mathbb{R}$ (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds). The other is the cut map $x \mapsto \{q \in \mathbb{Q} : q < x\}$, injective because $\mathbb{Q}$ is dense in $\mathbb{R}$ (ℚ is dense in every Archimedean ordered field), and $\mathcal{P}(\mathbb{Q})$ is a copy of $\mathcal{P}(\mathbb{N})$ because $\mathbb{Q}$ is countable ($\mathbb{Q}$ is countably infinite). The Schröder-Bernstein theorem closes the loop, and it is choice free, which is what makes clause (a) a theorem of ZF.

Why binary expansions are avoided. The textbook proof identifies a real in $[0,1]$ with the set of positions where its binary expansion has a $1$, and then has to deal with the reals having two expansions. Neither injection above meets that difficulty: the Cantor set map is already published as a bijection, and the cut map needs only density. The cost is that the two injections go in opposite directions and The Schröder-Bernstein theorem is required to combine them, which is free, since that theorem is choice free.

Which half needs the Axiom of Choice, and why. Clause (a) is an equinumerosity statement and is a theorem of ZF. Clause (b) writes $\lvert \mathbb{R}\rvert$ and $2^{\aleph_0}$ as cardinals, that is as ordinals, and in ZF alone $\mathcal{P}(\mathbb{N})$ need not be well-orderable, so those symbols need not denote anything. The hypothesis buys the notation, not the mathematics.

What is still not decided. Nothing here says which aleph $2^{\aleph_0}$ is. The one constraint proved in this development is that its cofinality is uncountable (Assuming the Axiom of Choice: $\kappa < \kappa^{\operatorname{cf}(\kappa)}$ for every infinite cardinal $\kappa$, and $\operatorname{cf}(2^{\kappa}) > \kappa$; in particular $\operatorname{cf}(2^{\aleph_0}) > \aleph_0$), and the question of whether it is $\aleph_1$ is the continuum hypothesis (What each result on this page costs in choice, and where the continuum escapes what ZFC can decide).