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
This is really just a special case of proof by cases:
(In your question, take $\psi$ to be $\perp$: a theory is inconsistent iff it proves $\perp$.)
Depending on your proof system, this may be included explicitly as one of the basic inference rules. (Alternatively, you might deduce it by first using the deduction theorem to rewrite your hypotheses as $\Gamma\vdash\varphi\rightarrow\psi$ and $\Gamma\vdash\neg\varphi\rightarrow\psi$.)