Hint
For Mendelson's system, see :
We need :
Lemma 1.8 [ page 27 ] : $\vdash \varphi \to \varphi$.
With it, (Ax.1) and (Ax.2), we can prove Prop.1.9 (Deduction Th) [ page 28 ] and some useful results [ page 29 ]:
Corollary 1.10(a) : $\varphi \to \psi, \psi \to \tau \vdash \varphi \to \tau$
and :
Lemma 1.11(b) : $\vdash \lnot \lnot \varphi \to \varphi$.
First, we can prove the "easy" version :
a) if $\Gamma \cup \{ \lnot \gamma \}$ is inconsistent, then $\Gamma ⊢ \gamma$.
Proof
1) $\Gamma \cup \{ \lnot \gamma \}$ is inconsistent, i.e. $\Gamma \cup \{ \lnot \gamma \} \vdash \varphi$ and $\Gamma \cup \{ \lnot \gamma \} \vdash \lnot \varphi$, for some formula $\varphi$
Thus :
2) $\Gamma \vdash \lnot \gamma \to \varphi$ --- from 1) by Ded.Th
3) $\Gamma \vdash \lnot \gamma \to \lnot \varphi$ --- from 1) by Ded.Th
4) $\vdash (\lnot \gamma \to \lnot \varphi) \to ((\lnot \gamma \to \varphi) \to \gamma)$ --- (Ax.3)
5) $\Gamma \vdash \gamma$ --- from 2), 3) and 4) by modus ponens twice.
Finally, for the sought result :
b) if $\Gamma \cup \{ \gamma \}$ is inconsistent, then $\Gamma ⊢ \lnot \gamma$,
we have to apply Noah's suggestion.
As in case a) above, we have :
1) $\Gamma \vdash \gamma \to \varphi$
2) $\Gamma \vdash \gamma \to \lnot \varphi$
3) $\vdash \lnot \lnot \gamma \to \gamma$ --- by Lemma 1.11(a)
4) $\Gamma \vdash \lnot \lnot \gamma \to \lnot \varphi$ --- from 2), 3) and Corollary 1.10(a)
5) $\Gamma \vdash \lnot \lnot \gamma \to \varphi$ --- from 1), 3) and Corollary 1.10
6) $\vdash (\lnot \lnot \gamma \to \lnot \varphi) \to ((\lnot \lnot \gamma \to \varphi) \to \lnot \gamma)$ --- (Ax.3)
7) $\Gamma \vdash \lnot \gamma$ --- from 4), 5) and 6) by modus ponens twice.
The only way a theory can be inconsistent is if it has no models. Because your observation is right: if a theory has a model $M$, then we can never have $M \models \phi$ and $M \models \neg \phi$ at the same time. So:
A theory $\Gamma$ is inconsistent if and only if $\Gamma$ has no models.
Now for your actual question: we have $\Gamma \not \models \neg \phi$. So there is a model $M$ of $\Gamma$, such that $M \not \models \neg \phi$. That is, $M \models \phi$. But then $M$ is a model of $\Gamma \cup \{\phi\}$, so we see that that theory is consistent.
Note that we did not need that $\Gamma \not \models \phi$. Indeed, if we would have $\Gamma \models \phi$ and $\Gamma$ is consistent, then $\phi$ is already true in every model so $\Gamma \cup \{\phi\}$ is definitely consistent (in fact, $\Gamma$ and $\Gamma \cup \{\phi\}$ are equivalent theories).
Best Answer
Your proof is entirely correct. Cheers :).