My attempt:
Proceed by contradiction, assume that G is soluble. Then every subgroup and quotient group of G is soluble, so in particular, the non-trivial perfect subgroup, which we call H, is soluble. But since H is perfect, the smallest normal subgroup (the derived subgroup H') of H where the quotient group H/H' is abelian, is H, so the subnormal series of H with abelian factors won't terminate, since we will have that H contains normal subgroup H' contains normal subgroup H'..... where none of the H'=H is = {e} since H is non-trivial.
Is this argument ok? I'm unsure if using the fact that the subnormal series doesn't terminate is strong enough for a contradiction.