proving that $P\ne NP$ under an assumption Suppose that $P^{SAT} \not\subseteq coNP$. Prove that $P\ne NP$.

What I did:

Suppose that $P=NP$. Then, $P = coP = NP = coNP$.

We know that $P^P = P$.

Then, by assumption: $P^{NP} = NP = coNP$

Since $SAT$ is $NP$-complete we get: $P^{SAT} = coNP$ in contradiction to the assumption.

So $P\ne NP$.

I've been told that my proof is false, but I don't understand why.
Here is a more direct proof. Suppose that $P=NP$. In particular, SAT can be solved in polynomial time. Therefore $P^{SAT} \subseteq P^P = P \subseteq coNP$.