Proving $\operatorname{card} \mathbb{R} = \operatorname{card} 2^{\mathbb{N}}$ *without* using Cantor-Schröder-Bernstein theorem? On math.stackexchange.com and elsewhere proofs of the equality $\operatorname{card} \mathbb{R} = \operatorname{card} 2^{\mathbb{N}}$, or equivalently the equality $\operatorname{card} \mathbb{R} = \operatorname{card} \mathcal{P}(\mathbb{N})$, abound that use the Cantor-Bernstein theorem.
What is a proof that does not use that theorem?
P.S. All the proofs that I previously knew, including those appearing in two undergraduate texts I authored, use CB and prove CB there, too.

This is my preferred proof, one that avoids the Cantor-Bernstein theorem. It is just a bit different from that in https://math.stackexchange.com/a/4193008/32337.
Denote $\mathbb{N} \setminus \{0\}$ by $\mathbb{N}^{\ast}$.
It suffices to show that  $\operatorname{card} 2^{\mathbb{N}^{\ast}} = \operatorname{card}(0, 1)$.
Let
$C$ be the set of binary sequences $(b_{n})_{n \in \mathbb{N}^{\ast}}$ that are eventually constant
and let $B = 2^{\mathbb{N}^{\ast}} \setminus C$,
the set of those that are not.
Since $B$ of $2^{\mathbb{N}^{\ast}}$ is denumerable
while the interval $(0, 1)$ is uncountable, a ``Hilbert's hotel'' maneuver shows that
$\operatorname{card} \bigl((0, 1) \cup B\bigr) = \operatorname{card} (0, 1)$.
Hence it suffices to show that $\operatorname{card} 2^{\mathbb{N}^{\ast}} = \operatorname{card} \bigl((0, 1) \cup B\bigr)$.
From order-completeness of $\mathbb{R}$,
for each $x \in (0, 1)$,
there is a unique $(x_{n})_{n \in \mathbb{N}^{\ast}} \in B$ for which
$x = \sum_{n =1}^{\infty} x_{n}/2^{n}$.
Define the map $f \colon 2^{\mathbb{N}^{\ast}} \to (0, 1) \cup B$ as follows:
If $b \notin C$, then $f(b) = \sum_{i=1}^{\infty} b_{i}/2^{i}$;
but if $b \in B$, then $f(b) = b$.
Then $f$ is the desired bijection.
