Let $n \ge 4$ and $u,v \in S_n$.
Prove that $uv$ is a commutator. In other words, prove that there are $\alpha, \beta \in S_n$ such that $uv = \alpha \beta \alpha^{-1} \beta^{-1}$.
This is the first part in an exercise which later asks to deduce that the derived group of $S_n$ is $A_n$ (the alternating group) for $n \ge 4$; it is easy to see that the above implies $A_n \subset S_n '$, and we have the other inclusion because $A_n \unlhd S_n$, and $S_n /A_n$ is Abelian.
Back to the question, I do not have any good ideas until now. If the transpositions are non-disjoint, and $n \ge 5$, then we might say: (where $u = (x \ \ y), v = (y \ \ z)$ and $a,b \in \{1,...,n\} \setminus \{x,y,z\}$):
$$uv = (x \ \ y \ \ z) = (a \ \ z \ \ x) (b \ \ z \ \ y)(a \ \ z \ \ x)^{-1} (b \ \ z \ \ y)^{-1} $$
And we are done.
Otherwise, I am stuck. Any suggestions?
In the non-disjoint case you should also consider what happens when $u=v$.
For the disjoint case, suppose $u=(x~y)$ and $v=(z~w)$ where $x,y,z,w$ are distinct. Notice that $u$ and $v$ are conjugates in $S_n$, so there exists $\alpha\in S_n$ such that $\alpha v\alpha^{-1}=u$ (what is $\alpha$ explicitly?). Then take $\beta=v$. Note that $v^{-1}=v$ since $v$ is a transposition. Then $$\alpha\beta\alpha^{-1}\beta^{-1}=(\alpha v \alpha^{-1})v^{-1}=uv.$$
If you have studied the conjugacy classes of $S_n$, you should be able to find $\alpha$. I can elaborate if needed.