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.
Dummit and Foote's Abstract Algebra text is definitely not lacking in terms of mathematical content. Although the comments above are correct when they say there is almost no category theory, the number of topics covered in the text makes up for it. I have the text in front of me and it totals 7 pages of category theory. On the other hand, I have found each main section of the text has a fairly good treatment of the respective topic. Here I am mainly talking about the Group Theory, Rings and Modules, and Field and Galois Theory sections. There are several chapters on more advanced topics at the end but from what I can tell those are pretty introductory. I would contend that if you wish to have a single textbook that is very comprehensive, relatively easy to follow, and all around good for learning Abstract Algebra, Dummit and Foote is the way to go.
If you already have a solid mathematical background though you may want to check out Serge Lang's Algebra, which is just as comprehensive but considered more of a "graduate level" text. It also includes more Category Theory if that is a factor.
Best Answer
I'll write out the answer in more detail.
Assume (*) $gNg^{-1} \subset N$ for all $g$.
Let $h \in N$ be arbitrary.
Now fix $g$. By (*) $g^{-1}hg \in N$. So then $h = g(g^{-1}hg)g^{-1} \in gNg^{-1}$. Since $h$ was arbitrary, this implies that $N \subset gNg^{-1}$, so in fact $N = gNg^{-1}$