Existence of a maximal subgroup I'm looking for a "fake" proof of the existence of a maximal subgroup using Zorn's lemma. Do either of you know where I might find this?

Here’s a trivial and obvious gappy proof:
Let $G$ be a nontrivial group. Let $P$ be the collection of all proper subgroups of $G$, partially ordered by inclusion. It is nonempty, since $\{e\}\in P$. If $\mathcal{C}$ is a nonempty chain in $P$, then let $K=\cup_{H\in\mathcal{C}}H$. Then $K$ is a subgroup: there exists $H\in \mathcal{C}$, and $e\in H$, so $e\in K$. If $x,y\in K$, then there exist $H_1,H_2\in \mathcal{C}$ with $x\in H_1$, $y\in H_2$. Since $\mathcal{C}$ is a chain, either $H_1\subseteq H_2$ or $H_2\subseteq H_1$. Either one, both $x$ and $y$ are elements of some $H_i\in\mathcal{C}$, hence $xy^{-1}\in H_i\subseteq K$. Thus, $K$ is a subgroup.
Herein lies the gap
By Zorn’s Lemma, $P$ has maximal elements. Let $M$ be a maximal element of $P$. If $M\leq N\leq G$, either $N=G$, or $N\neq G$. If $N\neq G$ then $N\in P$, so maximality of $M$ and $M\leq N$ gives $M=N$. Thus, $M$ is a maximal subgroup of $G$. “QED”
