Let $K$ be a minimal normal subgroup of $G$; $|K|$ is either $2$, $3$, $4$, or $5$.
- If $K$ has order $3$, then $G/K$ has order $20$, and has a normal Sylow 5-subgroup (by Sylow's theorems). The preimage is a normal subgroup of order $15$ in $G$. Again, by Sylow's theorems, the Sylow 5-subgroup is normal (in the subgroup, and then the whole group).
- If $K$ has order $4$, $G/K$ has order $15$, and the arguments above are basically reversed. Namely, $G/K$ has a normal Sylow 5-subgroup, we take the preimage, and get a normal subgroup of order $5$ again.
- If $K$ has order $5$, we're done.
What remains is the case $K$ has order $2$, and thus $G/K$ has order $30$. If the Sylow 5-subgroup of $G/K$ is normal, then just like above, $G$ has a normal subgroup of order $5$ (since the Sylow 5-subgroup of a group of order $10$ is always normal).
That means all that is left is to prove a group $H$ of order $30$ always has a normal Sylow 5-subgroup. We can proceed the same way we did above: find a minimal normal subgroup, look at the quotient, etc. The same arguments show $H$ has a normal Sylow 5-subgroup. [Thanks to @DerekHolt for mentioning this approach, which is pedagogically cleaner.]
[Original Version: This can be done by considering the action of $H$ on itself, by right multiplication. An element of order $2$ will act as an odd permutation, which shows $H$ has a normal subgroup of order $15$. Once more, we finish by noting that a group of order $15$ always has a normal Sylow 5-subgroup.]
Here's a second approach, which is more case-by-case, but also more elementary:
Assume $G$ has no normal Sylow 5-subgroup. Then it has $6$ such subgroups, and a case-by-case analysis shows it has $10$ Sylow 3-subgroups. Counting elements then shows $G$ has $5$ Sylow 2-subgroups. The conjugation action on these Sylow 2-subgroups embeds $G$ in $S_5$, which means $G\cong A_5$, contradicting solvability.