That seems largely right, except you’re conflating the Axiom of Choice with the Generalized Continuum Hypothesis at points.

The main effect of the Axiom of Choice for cardinality questions is that it tells us A) the notions “X has an injection into Y” and “Y has a partial surjection onto X” are the same, clearing up an ambiguity as to what “Y is at least as large as X” should mean, and B) it also ensures us that of any two different cardinalities, one is larger than the other, which makes these questions much simpler.

[This is distinct from the Generalized Continuum Hypothesis which is that the beth numbers (where beth_n is the powerset of beth_{n - 1}) match the aleph numbers (where aleph_n is the smallest (well-orderable) cardinal larger than aleph_{n - 1}), which makes things even simpler; the GCH implies the Axiom of Choice, but not vice versa]

Now, let’s consider V = {the set of all subsets of S with size at most K}, where S itself has size 2^K, and K is infinite.

Note that V is certainly at least as large as S (since the singleton operator injects S into V).

With the axiom of choice (but without requiring the GCH), we can get the converse as well: Note that W = {the set of all subsets of S with size at most K, along with a well-order on S} is at least as large as V [since every set can be given at least one well-order]. But W = S^0 + S^1 + S^2 + … + S^K <= S^K + S^K + S^K + … + S^K [with as many terms as there are ordinals <= K] <= 2^K * S^K [since the ordinals <= K can be identified with those particular subsets of K which are downwards closed, on some fixed well-ordering of K] = 2^K * (2^K)^K = 2^(K(K + 1)) = 2^K [since any infinite K satisfies K(K + 1) = K^2 = K] = S. Thus, S is at least as large as V as well, making the two equal in size.

Actually, we don’t need the full Axiom of Choice for this to go through; we simply need to assume that K is well-orderable.

Even if K is not well-orderable, we still know that S <= V <= 2^S, trivially. However, there’s no reason I can see why V should be injectable into S. I’ll try to provide proof that this can fail (for non-well-ordered K) later.