Question about maximal consistency 
Let $\sigma$ be a consistent set of propositions such that for every set $\gamma$, either $\sigma$ is proofwise stronger than $\gamma$ that is {$\alpha : \sigma \vdash \alpha$} $\supseteq$ {$\alpha : \gamma \vdash \alpha$} or $\sigma \cup \gamma$ is not consistent. Prove that in that case $\sigma$ is maximally consistent.

My proof:
Suppose $\sigma$ is not maximally consistent then there exist a proposition $\alpha$ such that $\sigma \not\vdash \alpha$ and $\sigma \not\vdash \neg\alpha$. In this case, let $\gamma$ = {$\alpha$}. Since $\sigma \not\vdash \alpha$ that means $\sigma$ is not proofwise stronger than $\gamma$, and $\sigma \cup \gamma$ is consistent (by Lemma below) , which contradicts our assumption.
Lemma: For every set of propositions Σ and every proposition α, if Σ $\not\vdash$ α
then Σ ∪ {(¬α)} is consistent.
Proof of the lemma: Assume, by way of contradiction, that Σ∪{(¬α)} is not
consistent. Then it proves every proposition, in particular Σ ∪ {(¬α)} $\vdash$ α. By
the deduction theorem we therefor get Σ $\vdash$ ((¬)α → α). However, $\vdash$(((¬)α →
α) → α), for every proposition α. Now, applying MP we get
Σ $\vdash$ α, which contradicts our initial assumption.
Is there a gap in my proof?

Yes, there's a gap in your proof because the assumption is not that $\sigma$ is proofwise stronger than $\gamma$, but that either $\sigma$ is proofwise stronger than $\gamma$ or $\sigma\cup\gamma$ is not consistent; so you have to address that second possibility, too.
