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.
One problem here is that the indexing is so unhelpful and confusing. Let's prove by induction on $i$ that $G^i \le Z_{c-i}(G)$, which is equivalent to the second containment you have to prove. It's true for $i=0$, since $G^0=Z_c(G)=G$. Assuming it's true for $i$, we get $G^{i+1} = [G,G^i] \le [G,Z_{c-i}(G)]$.
But $Z_{c-i}(G)/Z_{c-i-1}(G) = Z(G/Z_{c-i-1}(G))$ implies that $[G,Z_{c-i}(G)] \le Z_{c-i-1}(G)$, which completes the inductive step.
I am afraid that the left hand inequality is not true in general! Let $G = D_{16} \times C_2$ be the direct product of a dihedral group of order $16$ and a cyclic group of order $2$. This is nilpotent of class $3$, but the direct factor $C_2$ of $G$ is contained in $Z_1(G) = Z(G)$, but not in $G^1 = [G,G]$.
Best Answer
You can do this directly, given your definition of nilpotent.
Let $U$ be a proper subgroup of $G$. Then there exists a largest $j$ such that $Z_j\subseteq U$. Since $U$ is proper, $j\lt r$. Then $Z_{j+1}\nsubseteq U$. Since $Z_{j+1}/Z_j\leq Z(G/Z_j)$, it follows that any element in $Z_{j+1}/Z_j$ centralizes $U/Z_j$. Thus, if $x\in Z_{j+1}$ and $u\in U$, then $xux^{-1}\in uZ_j\subseteq U$. Hence $Z_{j+1}$ normalizes $U$, so the normalizer of $U$ contains elements not in $U$ (namely, elements of $Z_{j+1}\setminus U$, which is nonempty since $Z_{j+1}$ is not contained in $U$).