$$\begin{align}
(1) & \alpha \rightarrow \beta && [\text{HYP}] \\
(2) & \beta \rightarrow \gamma && [\text{HYP}] \\
(3) & \alpha && [\text{HYP}] \\
(4) & \beta && [\text{MP}(1,3)] \\
(5) & \gamma && [\text{MP}(2,4)] \\
(6) & \alpha \rightarrow \gamma && [\rightarrow\text{-intro}(3,5)] \\
(7) & (\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma) && [\rightarrow\text{-intro}(2,6)] \\
(8) & (\alpha \rightarrow \beta) \rightarrow ((\beta \rightarrow \gamma) \rightarrow (\alpha \rightarrow \gamma)) && [\rightarrow\text{-intro}(1,7)] \\
\end{align}$$
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.
Best Answer
Actual answer
First of all, no, there's no obligation to use everything in $\Phi$. The set $\Phi$ is the set of sentences you're allowed to use in a proof. Adding stuff to $\Phi$ only makes it easier to prove things.
You're also mixing up $\Phi$ and the set of all sentences in the language. Obviously $\Phi$ proves every sentence in $\Phi$. What you want to talk about is the sentences in general which $\Phi$ proves.
Now, on to the issue of inconsistency. There are two notions of inconsistency of a set of sentences $\Phi$:
$\Phi$ proves everything.
For some $p$, $\Phi$ proves both $p$ and $\neg p$.
Conveniently, in classical propositional logic these are equivalent: clearly if $\Phi$ is inconsistent in the first sense, it's inconsistent in the second sense, and conversely if $\Phi$ proves both $p$ and $\neg p$ we can use proof by contradiction to prove whatever we want from $\Phi$: "Suppose $\neg q$. Then ... we conclude $p$ and $\neg p$. So since we got a contradiction from $\neg q$, we have $q$."
So all you need to do is find a single sentence such that $\Phi$ proves both the sentence and its negation.
Irrelevant but interesting note
Now note that I said "classical logic" above. There are other logics out there, and in some of them it is not the case that proving a contradiction means proving everything. For such logics we do need to distinguish between the notions of inconsistency given above: generally, the second (weaker) one is called "inconsistent" while the first is called "trivial." In these logics, and unlike classical logic, inconsistent theories may be interesting; see e.g. here.