If $\pi$ is a set of primes, define $O_{\pi}(G)$ to be the subgroup of $G$ generated by all the normal $\pi$-subgroups of $G$.
Show that $O_{\pi}(G)$ is the intersection of all maximal $\pi$-subgroups of $G$.
Let $H$ to be the intersection of all maximal $\pi$-subgroups of $G$.
Let $\sigma\in Aut(G)$.
Then $\sigma$ permutes the maximal $\pi$-subgroups.
Hence $\sigma(H)= H$ and $H$ char $G$.
This means that $H\lhd G$ and by definition $H\leq O_{\pi}(G)$.
Now I need to show $O_{\pi}(G)\le H$.
I guess the idea will be show that every normal $\pi$-subgroup is contained in every maximal $\pi$-subgroup but I can't see why it is true.
I can't see anything wrong within your proof that $H$ is a characteristic subgroup and therefore a normal subgroup. And, $H$ is also a $\pi$-subgroup because every element in $H_i$ has order a $\pi$-number, so does every element in $H$. So why don't they define $O_{\pi}(G)$ as the maximal normal $\pi$-subgroup?
For your part, I want to prove by saying that every normal $\pi$-subgroup $N$ is a subgroup of every maximal $\pi$-subgroup $H_i$, because otherwise (at least in finite case), $NH_i$ is a larger $\pi$-subgroup (this is because $|NH_i|=\frac{|N||H_i|}{|N \cap H_i|}$ is a $\pi$-number, as Cauchy's theorem tells us that $|N|,|H_i|$ are $\pi$-numbers). So every generator of the group is a subgroup of the intersection. So the group $O_{\pi}(G)$ generated by them is also a subgroup of the intersection.
Not sure whether I am correct. Hope anyone can share their thoughts.