No. The proof contains all sorts of badness. In any particularly useful definition of "provable" (i.e. provable by a consistent system with a recursively enumerable set of axioms), Godel's theorem does indeed hold. The fact that the proof doesn't address exactly what it means by "provable" is another issue altogether - and a slightly worse one is that it doesn't say what it means by "property".
Let me address two issues. The first is that what this proof actually claims is:
If $P(n)$ is unprovable, then $P(n)$ is true.
However, in order to extend this to $P(n)$ is provable, we would need to prove that $P(n)$ is unprovable, not just take it as an assumption. This proof does not suppose that we can prove $P(n)$ is unprovable - merely that it is. If we cannot prove $P(n)$ is unprovable, then we cannot prove it is true by the logic of the proof. It is actually a necessary consequence of Godel's theorem that there are statements which are unprovably unprovable (see here). The proof does nothing to fix this hole.
A much worse issue is that the proof assumes that it is "easy" to verify any counterexample. The proof is introducing another assumption in thinking that a counterexample would provably be a counterexample - really, the statement it proves is:
If $P(n)$ is unprovable, but one could construct a proof of $\neg P(n)$ for any counterexample $n$, then $P(n)$ is true.
We could consider something like the negation of Fermat's Last Theorem - in particular let $P(n)$ be the statement that there are solutions to $x^{n}+y^n=z^n$. To prove $P(n)$ was false for some specific $n$, we would need to prove that there were no solutions - and there's no formulaic way to do that. Sure, we can get away with setting $P(n)$ to statements like "this turing machine halts in $n$ steps", where we can run it for $n$ steps to verify or disprove the statement (i.e. a proof of $\neg P(n)$ exists if it is true), but this is far from every statement we might be interested in.