On $NP=\Sigma_2^P$ from non-deterministic time?
Here is a simpler "proof" using the same ideas. I will show that NP=coNP. Indeed, suppose that $L \in \mathsf{coNP}$, and consider the machine which on input $x$, invokes the NP-oracle $\overline{L}$, and outputs the opposite value.

What goes wrong in this argument? Let us use the "witness" definition of NP: a language $L$ is in NP if there exist a polynomial $p$ and a polytime predicate $P$ such that $$ x \in L \Longleftrightarrow \exists |y| \leq p(|x|) \, P(x,y). $$
Now suppose that $L$ is in coNP, that is,
$$ x \in L \Longleftrightarrow \forall |y| \leq p(|x|) \, \lnot P(x,y). $$
(We get this by applying the previous definition to $\overline{L}$.)

Our machine deciding $L$ nondeterministically accepts as input $x,y$ and outputs $\lnot P(x,y)$. However, this accepts a different language $L'$ given by
$$ x \in L' \Longleftrightarrow \exists |y| \leq p(|x|) \, \lnot P(x,y). $$
The difference is the quantification &mdash; existential instead of universal.



Informally, this shows that nondeterministic machines cannot simulate nondeterministic oracles. However, they can simulate deterministic oracles: $\mathsf{NP}^L = \mathsf{NP}$ for all polytime $L$. This is because an NP-machine can simulate the computation of $L$ on any given input directly. The same cannot be said about nondeterministic oracles.