Prove that all cycles of $S_n$ are generated by the transpositions $\{(ij) \mid 1 \leq i < j \leq n\}$ (do not assume any other set of transpositions generate the cycles of $S_n$).
So if we have a cycle $(x_1 x_2 x_3 \ldots x_n)$ then we have to show that the transpositions as stated above: $(x_1 x_n)(x_1 x_{n-1}) \ldots (x_1 x_2) (x_2 x_n) (x_2 x_{n-1}) \ldots (x_2 x_3) \ldots (x_{n- 1} x_n)$ generate this cycle.
So if $k \neq 1 \neq n$, then $x_k$ is first ocurred in the cycle $(x_k, x_{k+1})$. So $x_k$ goes to $x_{k+1}$. Then $x_{k+1}$ goes to $x_{k-1}$ since the next cycle where $x_{k+1}$ is seen is in $(x_{k-1} x_{k+1}$), and then $x_{k-1}$ goes to $x_{k-2}$ (in $(x_{k-2} x_{k-1})$), etc. until finally $(x_2 x_3)$. Then $(x_1 x_2)$ and lastly $(x_1 x_3)$. So $x_k$ goes to $x_3$, which doesn't really help my case here.
The symmetric group is even generated by all adjacent transpositions, hence in particular by all transpositions $(ij)$ with $i<j$. Furthermore, in the notes of Keith Conrad you can find all proofs for many different sets of generators of $S_n$, in particular, Theorem $2.1$. The proof is just this: every $n$-cycle is the product of transpositions as follows: $$ (x_1\,x_2\,\ldots x_n)=(x_1\,x_2)(x_2\,x_3)\cdots (x_{n-1}\,x_n). $$ For, say, $(13526)$ this gives $(13)(35)(52)(26)$. Hence these transpositions (and so all transpositions) generate all cycles.