Let $A$ denote a set and $P(A)$ be the power set. By definition for cardinalities $|A|\le|B|$ iff there exists an injection $A \hookrightarrow B$. Note that there is an obvious surjection $P(A) \to A$.
Without the axiom of choice now can there also be a surjection $A \to P(A)$?
No. Cantor's theorem does not depend on the axiom of choice.
The proof is really constructive from this point of view. Given a function from $A$ to $\mathcal P(A)$ we construct a set which is not in its image.