The following proof is given in Kurzweil and Stellmacher:
Let $G\neq 1$ be a $\pi$-separable group such that every $\pi$-section or every $\pi'$-section of $G$ is solvable. Then the $\pi$-Sylow theorem holds in $G$.
Proof: Let $U,H\leq G$ such that $U$ is a $\pi$-group and $H$ is a Hall $\pi$-group. It suffices to show that $U$ is contained in a conjugate of $H$. We prove this by induction on $|G|$.
Let $1\neq N$ be a normal subgroup of $G$ and $\overline G=G/N$. By induction, there exists $\overline g\in \overline G$ such that $\overline U^{\overline g}\leq \overline H$. In turn, $U^gN=(UN)^g\leq HN$. Replacing $U$ be any conjugate, we may assume $U\leq HN$.
If $O_{\pi}(G)\neq 1$, then we can choose $N=O_{\pi}(G)$. In this case, $U\leq H$. If $O_{\pi}(G)\neq 1$, then let $N=O_{\pi'}(G)$. The $\pi$-subgroup $U$ is a complement of $N$ in $NU$. Since $H\cap NU$ is also such a complement (1.1.11), the theorem of Schur-Zassenhaus shows that $U$ is a conjugate to $H\cap NU\leq H$.
The result 1.1.11 quoted above is the following:
Let $G=UV$ for $U,V\leq G$. Then $H=U(V\cap H)$ for all $U\leq H\leq G$.
I don't understand how the above is applied. According to the claim, we have $NU=N(H\cap NU)$ and to apply 1.1.11 we would need to show $N\leq H$, which is false.
We have $G=HN$. Then $NU$ certainly contains $N$, so by 1.1.11 we have $NU=N(H\cap NU)$. So we apply 1.1.11 with $HN$ in the role of $G$, $N$ in the role of $U$, $H$ in the role of $V$, and $NU$ in the role of $H$. Alternatively, $N(H\cap NU)=NH\cap NU=NU$ by the modular law.