Suppose $\mathcal{P}$ is a property of groups such that if a group $G$ has $\mathcal{P}$, then all subgroups of $G$ have $\mathcal{P}$. Given that $G$ is poly-$\mathcal{P}$, I'm trying to show that every subgroup $H \leq G$ is poly-$\mathcal{P}$.
Since $G$ is poly-$\mathcal{P}$, we know that there exists a series of subgroups $$1 = G_0 \vartriangleleft G_1 \vartriangleleft \ldots \vartriangleleft G_n = G$$ such that all factors $\frac{G_{i+1}}{G_i}$ have $\mathcal{P}$.
My idea was to consider the series $$1 = H_0 \vartriangleleft H_1 \vartriangleleft \ldots \vartriangleleft H_n = H$$ where $H_i = H \cap G_i$. I've tried to show that $\frac{H_{i+1}}{H_i}$ can be written as a subgroup of $\frac{G_{i+1}}{G_i}$ by using the second isomorphism theorem: we know that $G_i \vartriangleleft G_{i+1}$ and $G_{i+1} \cap H \leq G_{i+1}$, since $G_{i+1}\cap H$ is a subgroup of $G$.
Hence we have the following equalities: $$\frac{G_{i+1}\cap H}{G_i \cap H} = \frac{G_{i+1}\cap H}{G_i \cap H \cap G_{i+1}} \cong \frac{G_i(G_{i+1}\cap H)}{G_i} \leq \frac{G_{i+1}}{G_i}$$ and since $\frac{G_{i+1}}{G_i}$ has $\mathcal{P}$, so does $\frac{G_i(G_{i+1}\cap H)}{G_i}$.
Is this argument valid?