= Solution
A <solvable group> has a finite <subnormal series> from the identity to the <group> with <abelian group> factors. An equivalent definition uses the <derived series>: set $G^{(0)}=G$ and $G^{(j+1)}=[G^{(j)},G^{(j)}]$; the series must reach the identity. For <completeness>, an abelian-factor series forces successive derived subgroups into its successive terms, while a terminating <derived series> itself supplies such a series.
If $H\leq G$, induction gives $H^{(j)}\leq G^{(j)}$: every <commutator> of elements of $H^{(j)}$ is also a <commutator> in $G^{(j)}$. Thus termination for $G$ implies termination for $H$, proving that every <subgroup> of a <solvable group> is solvable.
In $S_4$, use
$$
\{1\}\triangleleft V_4\triangleleft A_4\triangleleft S_4,
\qquad V_4=\{1,(12)(34),(13)(24),(14)(23)\}.
$$
The <Klein four-group> is invariant under <conjugation> in $S_4$; the other normality assertions follow from the <alternating group> being the kernel of the sign map. The <quotient group> factors are $V_4,C_3,C_2$, all abelian. Hence \b[$S_4$ is solvable].
For the last example take $A_5\times C_2$. Its <subgroup> $\{1\}\times C_2$ is nontrivial, proper and normal, so the product is not a <simple group>. Projection onto $A_5$ is surjective, and every quotient of a <solvable group> is solvable, because a homomorphism carries each <derived subgroup> onto the <derived subgroup> of its image. But the nonabelian <simple group> $A_5$ has $A_5'=A_5$: its <derived subgroup> is nontrivial and normal. Thus the product is \b[not solvable either]. One standard justification of the simplicity used here is that the nonidentity <conjugacy class> sizes of $A_5$ are $15,20,12,12$; no proper nontrivial <union> with the identity has size dividing $60$, as a <normal subgroup> would require by <Lagrange's theorem for finite groups>.
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