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.
Are you assuming your group $G$ is finite? In this case we can proceed by induction on $|G|$. You should add in your proof "Since $G$ is solvable, there is a [normal] subgroup $H$ such that both $H$ and $G/H$ are solvable [and $1<|H|<|G|$]." You have three cases:
-- if $H\subseteq K$, by inductive hypothesis, $K/H$ is a solvable subgroup of $G/H$ (where $|G/H|<|G|$ since $1<|H|$), and the fact that both $H$ and $K/H$ are solvable gives $K$ solvable;
-- if, $K\subseteq H$, then by inductive hypothesis $K$ is solvable (use $|H|<|G|$);
-- in the remaining case, you have $|K\cap H|<|H|$, so, by inductive hypothesis $K\cap H$ is solvable. Furthermore, $K/(K\cap H)\cong KH/H$ is a subgroup of $G/H$ and it is therefore solvable (use that $|G/H|<|G|$).
Best Answer
Since derived series cannot be used, I assume that an equivalent definition of a solvable subgroup is used: a group $G$ is called solvable if it has a normal series $1 = G_0 < G_1 \dots < G_n = G$ (that is, all subgroups $G_i$ are normal in $G$) such that all factor groups $G_i/G_{i-1}$ are abelian, $i = 1, 2,\dots, n$. Let now $H$ be a nontrivial normal subgroup in $G$. Let $i$ be the maximal possible integer such that $H \cap G_i = 1$. Notice that $i < n$ since $H$ is non-trivial. Then $A= H \cap G_{i+1}$ is a nontrivial subgroup, and, being the intersection of two normal subgroups, it is normal in $G$. Also, the image of $A$ in the abelian factor group $G_{i+1}/G_i$ is isomorphic to $A/(A\cap G_i) \simeq A/1 \simeq A$. Hence $A$ is isomorphic to a subgroup of an abelian group $G_{i+1}/G_i$ and is therefore abelian.
My answer would perhaps be more precise and useful if I knew what specific definition (of several equivalent definitions used) of solvable groups is used in your question.