Gödel's paradox: Why is "a proof that some universal statement is unprovable" not a valid proof that this statement is true? [duplicate]

The key step in your argument is when you say that if there is a counterexample to $\forall x.P(x)$, then the theory $T$ we speak about can prove that it is in fact a counterexample.

The particular form of $P(x)$ that comes out of Gödel's construction has the property that whenever we have a particular numeral $\bar n$, then the formula $P(\bar n)$ will always be true, and $T$ proves it to be true.

However, that does not constitute a proof of $\forall x.P(x)$. To think it does invokes a hidden assumption that everything our $\forall x$ ranges over can be expressed as a numeral. For sure this is true about our intended interpretation of $T$ as the "actual natural numbers", but in order for $T$ to prove something it needs not only to be true in the intended interpretation, but also in every other interpretation that satisfies the axioms of $T$.

Here the reasoning runs into trouble, because we know that $T$ must have some models where some of the elements don't correspond to numerals. (This is so independently of the incompleteness theorem; the "usual compactness argument" shows that such a model must exist for any theory that can speak about the naturals). Therefore, the known fact that $P$ is true and provable about all numerals doesn't imply that $\forall x.P(x)$ is true in all models, and therefore we can't conclude that it ought to be provable. And so the entire argument falls apart.

Gödel's notion of an "$\omega$-consistent" theory requires that there is no formula $\varphi(x)$ such that $T$ both proves $\varphi(\bar n)$ about every numeral, and proves $\exists x.\neg\varphi(x)$ which is the same as $\neg \forall x.\varphi(x)$. One way to show that a $T$ is $\omega$-consistent is to show that it has some model where all elements are numerals; interpreted in such a model the Gödel sentence will be true. The original incompleteness proof explicitly assumed that $T$ is $\omega$-consistent and used that to argue that $T$ cannot prove $\neg\forall x.P(x)$. Later, Rosser found a way to avoid this assumption and instead just assume that $T$ is consistent.

The thing you're missing is that "true" and "false" are always relative to a model, but in arithmetic they are relative to the standard model, unless stated otherwise.

Now, if there is a counterexample to a $\Pi_1$ statement, then it is a witness to its negation, a $\Sigma_1$ statement. But here's the thing, $\sf PA$ is $\Sigma_1$-complete: every $\Sigma_1$ statement which is true in $\Bbb N$ is in fact provable from $\sf PA$.

Now, if $\forall x\varphi(x)$ is a $\Pi_1$ statement which is neither provable nor disprovable from $\sf PA$, that means that it is necessarily the case that $\exists x\lnot\varphi(x)$ is false (in $\Bbb N$), otherwise, it would be provable. Therefore $\forall x\varphi(x)$ is true (in $\Bbb N$).