Subgroup of Solvable Group is Solvable/Proof 2

Proof
Firstly we, know that a group is solvable if and only if its derived series

$D(G)=[G,G] \, \ D^i(G)= [D^{i-1}(G),D^{i-1}(G)$

becomes trivial after finite iteration. Meaning

$D^j(G) = \{1\}$

for some finite j. Now it is trivial that

$D(H) \leq D(G)$

since H is smaller than G. Further since $D^i(H)$ is dominated by $D^i(G)$ it too has to become trivial after a finite amount of steps.