> Addressing your issue directly, the Axiom of Choice is actively debated:
The axiom of choice is not required to prove Cantor’s theorem, that any set has strictly smaller cardinality than its powerset.
Actually, I can recount the proof here: Suppose there is an injection f: Powerset(A) ↪ A from the powerset of a set A to the set A. Now consider the set S = {x ∈ A | ∃ s ⊆ A, f(s) = x and x ∉ s}, i.e. the subset of A that is both mapped to by f and not included in the set that maps to it. We know that f(S) ∉ S: suppose f(S) ∈ S, then we would have existence of an s ⊆ A such that f(s) = f(S) and f(S) ∉ s; by injectivity, of course s = S and therefore f(S) ∉ S, which contradicts our premise. However, we can now easily prove that there exists an s ⊆ A satisfying f(s) = f(S) and f(S) ∉ s (of course, by setting s = S), thereby showing that f(S) ∈ S, a contradiction.
Perhaps this is an ignorant question, but wouldn't you need AC to select the s ⊆ A whose existence the contradiction depends on? A constructive proof, at least the ones I'm trying to build in my head, stumbles when needing to produce that s to use in the following arguments.
No, because you only have to choose _one_ s for the proof to work, and a finite number of choices is valid in intuitionistic and constructive mathematics.
The axiom of choice is not required to prove Cantor’s theorem, that any set has strictly smaller cardinality than its powerset.
Actually, I can recount the proof here: Suppose there is an injection f: Powerset(A) ↪ A from the powerset of a set A to the set A. Now consider the set S = {x ∈ A | ∃ s ⊆ A, f(s) = x and x ∉ s}, i.e. the subset of A that is both mapped to by f and not included in the set that maps to it. We know that f(S) ∉ S: suppose f(S) ∈ S, then we would have existence of an s ⊆ A such that f(s) = f(S) and f(S) ∉ s; by injectivity, of course s = S and therefore f(S) ∉ S, which contradicts our premise. However, we can now easily prove that there exists an s ⊆ A satisfying f(s) = f(S) and f(S) ∉ s (of course, by setting s = S), thereby showing that f(S) ∈ S, a contradiction.