Let $\operatorname{Card}(X)$ denote the cardinal number of the set $X$. The standard proof of Cantor's Power Set theorem stating that "$\operatorname{Card}(X) < \operatorname{Card}(2^X)$" is simple, straightforward and does not use the Axiom of Choice.
If $X$ is any infinite set, is there a simple, straightforward proof of the statement "$2\operatorname{Card}(X) < \operatorname{Card}(2^X)$" which does not use the Axiom of Choice?
Simple. $2\cdot\operatorname{Card}(X)$ is the cardinality of $X\times\{0,1\}$.
Now consider the following injection: $$\langle x,i\rangle\mapsto\begin{cases}\{x\} & i=0\\ X\setminus\{x\} & i=1\end{cases}$$
This gives us $\leq$; whereas in the other direction, if $\operatorname{Card}(2^X)=2\cdot\operatorname{Card}(X)$, we have that $\operatorname{Card}(2^X)\leq \operatorname{Card}(X)^2$ (because it is always true that $2\cdot\operatorname{Card}(X)\leq\operatorname{Card}(X)^2$.
This is impossible, though, since the latter is false for any $X$ with at least five elements, let alone an infinite set.
To see a proof of the latter without choice, see Jech The Axiom of Choice, Lemma 11.10 (p. 160 of the Dover reprint).