Here's a hint - I don't know what you already know, so if you don't understand, just ask for clarification!
OK, so we're looking for two solvable subgroups $H$ and $K$ of a non-solvable group $G$, such that $HK = G$. The smallest non-solvable subgroup is indeed $A_5$, and every smaller group is solvable. In particular, $A_4$ and $C_5$ are solvable.
Can you find two subgroups of $A_5$, one isomorphic to $A_4$ and another isomorphic to $C_5$ which together generate $A_5$? Can you then show that not only $\langle H,K \rangle = G$ but also $HK = G$?
EDIT: In the previous wrong version I was hinting towards two copies of $A_4$ which together generate $A_5$, but do not have $HK = G$.
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.
Best Answer
Here's a hint:
If two subgroups $H$ and $K$ are normal and intersect precisely in $\{e\}$, then $HK = KH = H \times K$. What is the intersection of a Sylow $p$-subgroup and a Sylow $q$-subgroup for different $p$ and $q$?