Gödel didn’t prove the incompleteness?

Question

Gödel’s proof considers an arbitrary system K containing natural number. The proof defines a relation Q(x,y) then considers ∀x(Q(x,p)) where p is a particular natural number. The proof shows that the hypothesis that ∀x(Q(x,p)) is K provable leads to contradiction, so ∀x(Q(x,p)) is not K provable. The proof proves then ∀n(├ Q(n,p)) that means Q(n,p) is K provable for all natural number n. The proof introduces a new ω-consistent claiming that for any property B(x), it’s not the case ∀n(├ B(n)) ⋀ (├ ¬∀x(B(x))), and assumes that K is ω-consistent. The proof proves that ¬∀x(Q(x,p)) is not K provable by that the assumption that it’s K provable, that is, ├ ¬∀x(Q(x,p)), is in contradiction with the assumption that K is ω-consistent as we have already ∀n(├ Q(n,p)). But the proof neglects the link between truth value and proof. Considering the propositional logic theorem B → B, the assumption of ∀x(Q(x,p)) in K would prove ∀x(Q(x,p)) in K then leads to contradiction, so we have ¬∀x(Q(x,p)), and ∀x(Q(x,p)) is not K provable as K is consistent, and then ∀n(├ Q(n,p)). In another side, the assumption of ¬∀x(Q(x,p)) in K would prove it in K and then lead to contradiction with the assumption that K is ω-consistent, so we have ∀x(Q(x,p)). Then, we have that either ∀x(Q(x,p)) is a paradox or K is not ω-consistent. But in both cases the proof doesn’t prove the incompleteness.

With this proof can we continue to consider that the incompleteness is proven?