With your definition, to show that if $G$ is solvable then $N$ is solvable, let $$ 1 =G_0 \triangleleft G_1\triangleleft\cdots\triangleleft G_{m-1}\triangleleft G_m=G$$ be such that $G_{i+1}/G_{i}$ is abelian for each $i$.

Note: We do not need to assume that $N$ is normal; the argument below works just as well for any subgroup of $G$, not merely normal ones.

Let $N_i = G_i\cap N$. Note that since $G_i\triangleleft G_{i+1}$, then $N_i\triangleleft N_{i+1}$: indeed, if $x\in N_i$ and $y\in N_{i+1}$, then $yxy^{-1}\in N$ (since $x,y\in N$) and $yxy^{-1}\in G_i$ (since $G_i\triangleleft G_{i+1}$), hence $yxy^{-1}\in N\cap G_i = N_i$.

So we have a sequence $$1 = N_0\triangleleft N_1\triangleleft\cdots\triangleleft N_{m} = N.$$ Thus, it suffices to show that $N_{i+1}/N_i$ is abelian for each $i$.

Note that $N_{i} = N\cap G_i = (N\cap G_{i+1})\cap G_i = N_{i+1}\cap G_i$.

Now we simply use the isomorphism theorems: $$\frac{N_{i+1}}{N_i} =\frac{N_{i+1}}{N_{i+1}\cap G_i} \cong \frac{N_{i+1}G_i}{G_i} \leq \frac{G_{i+1}}{G_i}$$ since $N_{i+1},G_i$ are both subgroups of $G_{i+1}$ and $G_i$ is normal in $G_{i+1}$, so $N_{i+1}G_i$ is a subgroup of $G_{i+1}$.

But $G_{i+1}/G_i$ is abelian by assumption, hence $N_{i+1}/N_i$ is (isomorphic to) a subgroup of an abelian group, hence also abelian, as desired.