Is my solution correct? subgroups and quotients of solvable groups are solvable

As Eric Auld mentioned in the comment, the statement that $G_0/N=1/N$ is incorrect. To fix this, you may use the normal chain $$\begin{equation} 1=G_0N/ N \triangleleft G_1N/ N \triangleleft \dots \triangleleft G_sN/ N =G/N. \end{equation}$$

To show subgroup of solvable group is solvable. .. Let H is subgroup of G where G is solvable $G^k=\{e\}$. Claim: $H^n=G ^n$ for all $n \geq 1$. We will prove the claim by applying induction.

For $n=1$

$H^1=[H, H] $ is a subgroup of $[G, G] =G^1$.

Suppose , result is true for $n-1$.

Now, $H^n=[H^{(n-1)}, H^{(n-1)}]$ is subgroup of $[G^{(n-1)}, G^{(n-1)}]=G^n$

In particular, $H^k$ is subgroup of $G^k=\{e\}$

$H^k={e}$ and so , $H$ is solvable