Finite non-nilpotent and every proper subgroup is nilpotent implies not simple

Statement

Suppose $G$ is a finite group that is not a nilpotent group, but every proper subgroup of $G$ is a nilpotent group (and in particular, a finite nilpotent group). Then, $G$ cannot be a simple group.

Facts used

1. Nilpotent implies normalizer condition: In a nilpotent group, every proper subgroup is properly contained in its normalizer.
2. Finite and any two maximal subgroups intersect trivially implies not simple non-abelian

Proof

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Given: A finite non-nilpotent group $G$ such that every proper subgroup of $G$ is nilpotent.

To prove: $G$ is not simple.

Proof: We assume that $G$ is simple, and derive a contradiction. Let $n$ be the number of elements of $G$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	The intersection of any two maximal subgroups of $G$ is contained in an intersection of two maximal subgroups that is a normal subgroup of $G$.	Fact (1)	Every proper subgroup of $G$ is nilpotent. $G$ is finite		[SHOW MORE] Start with $K$, an intersection of two maximal subgroups of $G$. Suppose $L$ and $M$ are maximal subgroups of $G$ such that $L\cap M$ is maximal among all intersections of two maximal subgroups that contain $K$, i.e., it is not properly contained in any other intersection of two maximal subgroups of $G$. Such a maximal intersection exists by the finiteness of $G$. Consider the subgroup $N_{L}(L\cap M)$. This contains $L\cap M$ strictly (by Fact (1) and the fact that $L$ is nilpotent on account of being proper in $G$) and is contained in $L$. However, it is not contained in any maximal subgroup of $G$ other than $L$ by the maximality assumption on $L\cap M$. Similarly, $N_{M}(L\cap M)$ is not contained in any maximal subgroup of $G$ other than $M$. Thus, $N_{G}(L\cap M)$, which contains both $N_{L}(L\cap M)$ and $N_{M}(L\cap M)$, is not contained in any maximal subgroup of $G$. Because of the finiteness of $G$, it must be the whole group $G$. Thus, $L\cap M$ is normal in $G$.
2	Any two maximal subgroups of $G$ intersect trivially.		$G$ is simple	Step (1)	[SHOW MORE] By Step (1), the intersection is contained in an intersection of two maximal subgroups wherein the latter intersection is normal. Moreover, the latter intersection is also a proper subgroup. Since $G$ is simple, the latter intersection must be the trivial subgroup, and therefore the original intersection must also be the trivial subgroup.
3	We have the desired contradiction.	Fact (2)	$G$ is simple non-nilpotent, hence simple non-abelian	Step (2)	Follows directly from Fact (2).