Prove that there doesn't exist any normal subgroup $H$ such that $S_5/H $ is isomorphic to $S_4$

You are on the right track, you've shown that such a normal subgroup would need to be generated by a $5$ cycle in $S_5$, so now we need to show that the subgroup generated by a $5$ cycle in $S_5$ can't be normal. Conjugating a $5$ cycle will always yield a $5$ cycle (since conjugation preserves the order of elements of a group), so we'll need to work a little harder. There are a few ways to see this, but I'll give a hands on proof. Its enough to show that from any $5$ cycle, we can find something in the subgroup generated by its conjugates that isn't a $5$ cycle, since then we can't have any group generated by a $5$ cycle being normal. So if we have $(abcde)$ is our $5$ cycle, conjugating by the transposition $(ab)$ yields the $5$ cycle $(bacde)$, and $(abcde)(bacde)^{-1}=(cba)$, which has order $3$, so gives something bigger than our hypothetical normal subgroup of order $5$. Thus, no normal subgroup of order $5$ can exist.

You have almost finished the proof. So $|H|=5$, $H$ consists of 4 5-cycles and the identity element. Since there are $24$ cycles of length $5$ in $S_5$ there exists a $5$-cycle $c$ not in $H$. Since any two $5$-cycles are conjugate $S_5$, there is a conjugation taking a cycle from $H$ to $c$. This is a contradiction since $H$ is normal.

You're doing well. Up to a relabeling of the elements which $S_5$ operates on, the generator of $H$ can be chosen to be $(12345)$. Now observe $$ (12)(12345)(12)=(13452)\notin H $$ and you're done.