A simple solvable group must be cyclic of prime order (since it must be abelian, and so cannot have proper [normal] subgroups). But a simple solvable group would not contain nontrivial normal subgroups, so the proposition would be true for such a group by vacuity (the hypotheses are never satisfied). Alternatively, if you allow the whole group to be a "nontrivial normal subgroup", your $H$ can only be $G$ itself, which is already abelian, so you can set $A=H=G$; either way, the proposition is true for such a group.
(If $G$ is solvable, then $[G,G]$ is a proper subgroup of $G$; since it is always normal in $G$, if $G$ is also simple, then we must have $[G,G]=\{1\}$, hence $G$ is abelian).
Hint for the question. If $H$ is abelian, you are done. If not, then $[H,H]$ is nontrivial; use the fact that $H^{(n)}\subseteq G^{(n)}$, where $G^{(k)}$ is the $k$th term of the derived series of $G$, to show that $H$ is solvable, and use the fact that $H\triangleleft G$ to show $[H,H]\triangleleft G$. Then replace $H$ with $[H,H]$, lather, rinse, and repeat.
Different hint. Let $1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_s = G$, with $G_i/G_{i+1}$ abelian. By Problem 8, you can pick the $G_i$ normal in $G$. Look at $H_i=H\cap G_i$.
Added. Sigh. I didn't notice that Problem 8 assumes $G$ is finite; the result is true, as witnessed by the derived series, but again it's probably not what you want.
Added 2. Okay, this should work; it takes some of the ideas of the hint in Problem 8 of the same page, so it should be "reasonable". Let $i$ be the largest index such that $G_i\cap H$ is a proper subgroup of $H$. Then $H\subseteq G_{i+1}$, hence $G_i\cap H\triangleleft H$. Moreover, $H/(H\cap G_i) \cong (HG_i)/G_i \leq G_{i+1}/G_i$, so $H/(H\cap G_i)$ is abelian. As in problem 8, this means that $x^{-1}y^{-1}xy\in H\cap G_i$ for all $x,y\in H$. Show that this is also true for all $G$-conjugates of $H\cap G_i$, hence their intersection, which is normal in $G$, contains all $x^{-1}y^{-1}xy$ with $x,y\in H$. If this is trivial, then $H$ is abelian and we are done. If it is not trivial, then this intersection is nontrivial, and normal in $G$. Replace $H$ with this intersection, and note that the largest index $j$ such that the intersection with $G_j$ is proper is striclty smaller than $i$; so you can set up a descent. Lather, rinse, and repeat.
From the discussion in the comments two possible solutions arised, here is the first:
Since the group is nilpotent, upper central series terminates. Then for some $n$, $Z_n(G) = G$ and there must be a smallest $i$ such that $N \cap Z_i(G) = \langle e \rangle$ but $N \cap Z_{i+1}(G) \neq \langle e \rangle$. The goal is to show that $N \cap Z_{i+1}(G) \subseteq Z_1(G) = Z(G)$.
I will denote $Z_i(G),Z_{i+1}(G)$ simply by $Z_i, Z_{i+1}$.
We notice that the subgroup $[G,N]$ generated by the elements $\{ gng^{-1}n^{-1}\mid g \in G, n \in N\}$ is contained in the normal subgroup $N$. This is because for every generator $gng^{-1}n^{-1}$ of $[G,N]$, $gng^{-1}n^{-1} = (gng^{-1})n^{-1} \in N$. So in conclusion $[G,N] \subseteq N$.
Further we notice that since $Z(G/Z_i) = Z_{i+1}/Z_i$, then for any $c \in Z_{i+1}, g \in G$ we have that $Z_igc = Z_icg$, which implies $Z_igcg^{-1}c^{-1} = Z_i$ and $gcg^{-1}c^{-1} \in Z_i$ for all $g \in G, c \in Z_{i+1}$. These are exactly the generators of the group $[G, Z_{i+1}]$, therefore $[G, Z_{i+1}] \subseteq Z_i$
Now we look again at the nontrivial group $N \cap Z_{i+1}$. We established earlier that $[G,N] \subseteq N$ and $[G, Z_{i+1}] \subseteq Z_i$. This gives $[G, N \cap Z_{i+1}] \subseteq [G,N] \cap [G,Z_{i+1}] \subseteq N \cap Z_i = \langle e\rangle$. So actually the subgroup $[G, N \cap Z_{i+1}]$ is trivial, which happens when $N \cap Z_{i+1} \subseteq Z_1 = Z(G)$. This gives that $N \cap Z(G) \neq \langle e \rangle$.
And a perhaps a bit simpler one, relying on an alternative characterisation of nilpotent groups:
By an exercise in my book (Hungerford's Algebra, chapter 2, section 7, exercise 4), a group is nilpotent if and only if the $\gamma_m(G) = \langle e \rangle$ for some $m$, where $\gamma_1(G)=G,\gamma_2(G)=[G,G]$ and $\gamma_i(G)=[\gamma_{i−1}(G),G]$.
Given this, we define a sequence $N_1(G) = N, N_2(G) = [N,G]$ and $N_i(G) = [N_{i-1}(G),G]$, where $N$ is any proper normal subgroup of $G$.
Obviously for any $i$, $N_i(G) \subset N$ by normality of $N$.
It is also clear that $N_1(G) = N \subset G = \gamma_1(G)$. Assume inductively that $N_i(G) \subset \gamma_i(G)$. By definition, $N_{i+1}(G) = [N_{i}(G),G]$ and $\gamma_{i+1}(G)=[\gamma_{i}(G),G]$. It is clear then that $N_{i+1}(G) \subset \gamma_{i+1}(G)$.
But since $G$ is nilpotent, for some $m$, $N_m(G) \subset \gamma_m(G) = \langle e \rangle$ So in particular we get that $[N_{m-1}(G), G] = \langle e \rangle$. This can only be if $N_{m-1}(G) \subset Z(G)$, and since $N_{m-1}(G) \subset N$, we have that $N \cap Z(G) \neq \langle e \rangle$.
Best Answer
Consider the upper central series of $G$, $Z^0(G)=1\leqslant Z(G)\leqslant \cdots \leqslant Z^c(G)=G$, for some $c$. There is a least $k$ such that $H\cap Z^k(G)\neq 1$. Normality gives $[H\cap K,G]\leqslant H\cap [K,G]$. Use this and $[Z^k(G),G]\leqslant Z^{k-1}(G)$ to conclude.