How to prove $C$ from $A \leftrightarrow (B \leftrightarrow C)$ and $A \leftrightarrow B$? How does one prove  $C$ from the premises: $A  \leftrightarrow (B  \leftrightarrow  C)$ and $A  \leftrightarrow  B$ ?
I've tried to prove $C$ by contradiction, using a sub-proof which presumes $\neg C $, but although I can conclude all of the following in the subproof: $\neg A$, $ \neg B$, $ \neg (B \leftrightarrow C)$, I'm unable to find a contradiction this way.
I've been stuck on this for the whole day, and I think I might be over-thinking the problem.
Note: I want to prove this using the basic first-order logic rules (I'm using the First-Order Logic from the Language, Proof and Logic book).
 A: This basically follows from the associativity of $\leftrightarrow$. But let's pretend that we didn't know that.

We consider two exhaustive, mutually exclusive cases.
Case 1: Suppose that $B$ is true. Then since $A \leftrightarrow B$ is true, we know that $A$ is true. Thus, since $A  \leftrightarrow (B  \leftrightarrow  C)$ is true, we know that $B  \leftrightarrow  C$ is true. But then since $B$ is true, we know that $C$ is true, as desired.
Case 2: Suppose that $B$ is false. Then since $A \leftrightarrow B$ is true, we know that $A$ is false. Thus, since $A  \leftrightarrow (B  \leftrightarrow  C)$ is true, we know that $B  \leftrightarrow  C$ is false. But then since $B$ is false, we know that $C$ is true (otherwise, if $C$ was actually false, then $B \leftrightarrow C$ would be true, a contradiction). So we're done!
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \text{"#2"} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
$Your proof by contradiction approach is fine, here is how you can complete your proof.
Assume $\;C\;$ is false, then
$$\calc
A  \leftrightarrow (B  \leftrightarrow  C)
\calcop{\leftrightarrow}{using what we know about $\;C\;$}
A  \leftrightarrow (B  \leftrightarrow  \text{false})
\calcop{\leftrightarrow}{left hand side: use $\;A \leftrightarrow B\;$; right hand side: simplify}
B  \leftrightarrow \lnot B
\calcop{\leftrightarrow}{logic}
\text{false}
\endcalc$$
which is a contradiction.  Therefore $\;C\;$ is true.
A: Due to the transitivity of $\leftrightarrow$ and due to the fact that $A$ comes up on both premises 'at the same level', I find it natural to focus on $A$ and let it act as a pivot of sorts.
Start by proving $A\lor \neg A$ and perform $\lor$-$\text{Elim}$ on this disjunction.
In the first case just use $\leftrightarrow$-$\text{Elim}$ successively on the premises to get $C$.
In the second case (where one starts a subproof with the premise $\neg A$), use the premise $A\leftrightarrow B$ to get $\neg B$  and the premise $A\leftrightarrow (B \leftrightarrow C)$ to get $\neg(B\leftrightarrow C)$ (in both cases by negation introduction).
Now assume $\neg C$, prove $\neg B\leftrightarrow \neg C$ and from this last statement get $B\leftrightarrow C$.
At this point you can find a contradiction allowing you to conclude $C$ in the subproof whose premise is $\neg A$.
I leave the proof below.



A: 
I've tried to prove $C$ by contradiction, using a sub-proof which presumes $\neg C $, but although I can conclude all of the following in the subproof: $\neg A$, $ \neg B$, $ \neg (B \leftrightarrow C)$, I'm unable to find a contradiction this way.

This is a wee bit late, but if a contradiction may be derived from assuming $\neg C$, then you should also be able to derive $\def\too{\leftrightarrow}(B\too C)$ too, and from that you can obviously derive $A$, $B$, and $C$ in turn.

$$\def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}}\fitch{~~1.~~A\too(B\too C)\\~~2.~~A\too B}{~~3.~~A\to(B\too C)\hspace{10ex}\too\!\mathsf e~1\\~~4.~~(B\too C)\to A\hspace{10ex}\too\!\mathsf e~1\\~~5.~~A\to B\hspace{17ex}\too\!\mathsf e~2\\~~6.~~B\to A\hspace{17ex}\too\!\mathsf e~2\\\fitch{~~7.~~\lnot C\hspace{18ex}\mathsf a}{\fitch{~~8.~~B\hspace{16ex}\mathsf a}{~~9.~~A\hspace{15ex}\to\!\mathsf e~8,6\\10.~~B\too C\hspace{10ex}\to\!\mathsf e~9,3\\11.~~B\to C\hspace{10ex}\too\!\mathsf e~10\\12.~~C\hspace{15.5ex}\to\!\mathsf e~8,12}\\13.~~B\to C\hspace{13.5ex}\to\!\mathsf i~8{-}12\\\fitch {14.~~C\hspace{16ex}\mathsf a}{15.~~\bot\hspace{16ex}\neg~\mathsf e\,14,17\\16.~~B\hspace{16ex}\mathsf x~15}\\17.~~C\to B\hspace{13ex}\to\!\mathsf i~14{-}16\\18.~~B\too C\hspace{13ex}\too\!\mathsf i~13,17\\19.~~A\hspace{18ex}\to\!\mathsf e~18,4\\20.~~B\hspace{18ex}\to\!\mathsf e~19,5\\21.~~C\hspace{18ex}\to\!\mathsf e~20,13\\22.~~\bot\hspace{18ex}\neg\,\mathsf e~21,7}\\23.~~\neg\neg C\hspace{18ex}\neg~\mathsf i~7{-}22\\24.~~C\hspace{21ex}\neg\neg\,\mathsf e~23}$$

A: I would use the algebraic machinery:
$A\leftrightarrow B\quad$ iff $\quad 1\oplus A\oplus B$
and get $(A\leftrightarrow (B\leftrightarrow C))\wedge (A\leftrightarrow B)\quad$ iff 
$\quad (1\oplus A\oplus 1\oplus B\oplus C)(1\oplus A\oplus B)=$
$=(A\oplus B\oplus C)(1\oplus A\oplus B)$
$=A\oplus B\oplus C\oplus A\oplus AB\oplus AC\oplus AB\oplus B\oplus BC=$
$=C\oplus AC\oplus BC=C(1\oplus A\oplus B)\quad$ iff $\quad C\wedge(A\leftrightarrow B)$, which imply C.
Which considering the note may not be as relevant, but...
