Weird "proof" recursion
I feel like there is a kind of circularity in the last part of your argument (and I think the conclusion is wrong, there are undecidable problems which we cannot prove to be undecidable).
You are basically making a case distinction:

if $\alpha$ can be proved or $\neg \alpha$ can be proved: Then $P(\alpha)$ can be proved and thus $P(P(\alpha))$ is true. (I agree here)
if neither $\alpha$ nor $\neg \alpha$ can be proved: Then $\neg P(\alpha)$ can be proved and thus $P(P(\alpha))$ is true. I disagree here. The correct assumption to make this conclusion would be: &quot;if we can prove that neither $\alpha$ nor $\neg \alpha$ can be proved&quot;.

Note that you cannot in general prove that you are in the second case. You say you can because otherwise you could prove that you are in the first case. But what you need is to be able to prove that you cannot find a proof that you are in the first case, which is a different thing.