Let $L$ be a Lie algebra.
(a) If $L$ is solvable, then so are all subalgebras and homomorphic images of $L$.
(b) If $I$ is a solvable ideal of $L$ such that $L/I$ is solvable, then $L$ itself is solvable.
Proof of (b): Say $(L/I)^{(n)}=0$. Applying part (a) to the canonical homomorphism $\pi:L\rightarrow L/I$, we get $\pi(L^{(n)})=0$, or $L^{(n)}\subset I=\operatorname{Ker}\pi$.
How does $(L/I)^{(n)}=0$ imply $\pi(L^{(n)})=0$? I don't see why this follows from part (a).
Because $$\pi(L^{(n)}) \subseteq (L/I)^{(n)}.$$
This statement is pretty easy to prove by induction, and a good exercise, so I leave it to you.