Every $\pi$-separable group contains Hall $\pi$-subgroups A group $G$ is $\pi$-separable if G has a subnormal series such that each factor is either a $\pi$-group or $\pi'$-group.
This is a proof found in the book 'Theory of Finite Groups' by Kurzweil-Stellmacher, but I am unable to follow the proof
Let $G \neq 1$ be a $\pi$-separable group and $N\neq 1$ be a normal subgroup of $G$. Since $G/N$ is  $\pi$-separable, we may assume by induction that $G/N$ contains a Hall $\pi$-subgroup $H/N$ where $N\leq H\leq G$
Now $O_\pi(G)$ is the largest normal $\pi$-subgroup of $G$. If $O_\pi(G) \neq 1$ then choose $N = O_\pi(G)$ so $H$ is a Hall $\pi$-subgroup of $G$
I'm not sure why they chose $N$ to be $O_\pi(G)$ and how does $H$ become a Hall $\pi$-subgroup of $G$?
 A: First observe this: let $G$ be a finite group and $\pi$ a set of primes. If $N \unlhd G$, then $O_{\pi}(N) \unlhd O_{\pi}(G)$. This follows from the fact that $O_{\pi}(N)$ char $N \unlhd G$, which implies that $O_{\pi}(N) \unlhd G$. But $O_{\pi}(N)$ is then a normal $\pi$-subgroup and $O_{\pi}(G)$ is the largest of the sort of $G$. 
Now this implies that for any subnormal subgroup $S$ of $G$ we have $O_{\pi}(S)$ is a subnormal subgroup of $O_{\pi}(G)$: if $S=S_0 \lhd S_1 \lhd \cdots \lhd S_{k-1} \lhd S_k=G$, then by the previous $O_{\pi}(S) \unlhd O_{\pi}(S_1) \unlhd \cdots \unlhd O_{\pi}(G)$.
These observations imply that in a non-trivial $\pi$-separable group $G$ either $O_{\pi}(G) \neq 1$ or $O_{\pi'}(G) \neq 1$: look at the second last member of the subnormal series of $G$, which is either a non-trivial $\pi$- or $\pi'$-group!
Now this helps with the induction proof of your statement: assume first that $O_{\pi}(G) \neq 1$. Then $G/O_{\pi}(G)$ has smaller order than $G$ and by induction this group has a Hall $\pi$-subgroup, say $H/O_{\pi}(G)$, and then $H$ is the sought Hall $\pi$-subgroup.
So assume that $O_{\pi}(G) =1$, which implies that $O_{\pi'}(G) \neq 1$. Then $G/O_{\pi'}(G)$ has smaller order than $G$ and by induction this group has a Hall $\pi$-subgroup, say $K/O_{\pi'}(G)$. Observe that $O_{\pi'}(G)$ is a normal Hall subgroup of $K$. So here we can apply the Schur-Zassenhaus Theorem: there exists a subgroup $H \lt K$, with $K=HO_{\pi'}(G)$ and $H \cap O_{\pi'}(G)=1$. But then $H$ is a Hall $\pi$-subgroup, since the indexes $|G:K|$, $|K:H|=|O_{\pi'}(G)|$ are all $\pi'$-numbers. 
