Let $G$ be nilpotent and nontrivial. Since every maximal subgroup of $G$ must be normal, and if $M$ is maximal then $G/M$ has no proper subgroups, it follows that if $M$ is maximal then $G/M$ is a group of order $p$, hence abelian. Therefore, $[G,G]\subseteq M$, since $[G,G]$ is contained in any normal subgroup $N$ of $G$ such that $G/N$ is abelian. In particular, $[G,G]\neq G$. Now simply note that being nilpotent is inherited to subgroups, as proven below, to conclude that $[H,H]\neq H$ for all subgroups $H$ of $G$ when $G$ is nilpotent. Hence, if every subgroup of $G$ is subnormal ($G$ is nilpotent), then the commutator subgroup of $H$ is properly contained in $H$ for any nontrivial subgroup $H$ of $G$ ($G$ is solvable).
(If $H\leq G$ and $K$ is a subgroup of $H$, then $K$ is subnormal in $G$, so there exist subgroups $K\triangleleft K_1\triangleleft K_2\triangleleft\cdots\triangleleft K_m=G$. Intersecting the subnormal series with $H$ gives you a subnormal series fo $K$ in $H$, showing $K$ is subnormal in $H$ as well; thus, every subgroup of $H$ is subnormal, so subgroup of nilpotent is nilpotent).
Added. I am tacitly assuming above that $G$ has maximal subgroups; so it might fail for infinite groups in which every subgroup is subnormal. In the infinite case, the usual definition of "nilpotent" is via either the upper central series or the lower central series, and that of "solvable" via the derived series. In the case of the definition via the lower central series, proving solvability is very easy: recall that the lower central series of $G$ is defined inductively by letting $G_1=G$ and $G_{n+1}=[G_n,G]$; and a group $G$ is nilpotent if and only if $G_{n+1}=\{1\}$ for some $n\geq 1$. Now note that $G^{(2)}=[G,G]=G_2$, and if $G^{(k)}\subseteq G_n$, then $G^{(k+1)} = [G^{(k)},G^{(k)}] \subseteq [G_k,G]=G_{k+1}$. So if the lower central series terminates, then so does the derived series, proving that if $G$ is nilpotent then $G$ is solvable.
The idea is to see that if $ L $ is a $p$-group such that $ |L| = p^i $ with $ i > 1 $, then it has nontrivial center, and therefore $ Z(L) $ is a nontrivial normal subgroup. Let $ H_i $ and $ H_{i+1} $ be in the normal series, then $ L = H_{i+1}/H_i $ has a nontrivial normal subgroup $ Z(L) $, and by the correspondence theorem this means that there is a normal subgroup of $ H_{i+1} $ properly containing $ H_i $. In this way, you can continue to "break up" the normal series until each quotient is of prime order.
If $ Z(L) = L $, then any subgroup of $ L $ is normal, so take any element which does not generate $ L $ and take the cyclic subgroup generated by that element.
Best Answer
Let $N$ be a maximal normal subgroup. Then $G/N$ is simple and abelian (because it is smaller than $A_5$), hence cyclic. $N$ is strictly smaller than $G$, hence we may assume by induction that itis solvable, hence $G$ is solvable.