Question on a step of "A Simple Proof of Zorn's Lemma" by Lewin I'm reading Jonathan Lewin's "A Simple Proof of Zorn's Lemma" and cannot see the justification for one statement (which I've labeled Lemma 3 below). I'll summarize the proof below (adding a few lemma numbers for ease of communication). 
Immediately after stating Lemma 2, Lewin claims, "It now follows easily that [Lemma 3]." It does not follow so easily for me. In particular, Lemma 3 requires verifying the two properties of a "conforming set" (see below), and I find neither one to follow obviously from Lemma 2. Please avoid using any results about ordinals and transfinite recursion/induction in the clarification, since the whole point of this "simple" proof is to omit those concepts.

Proof Summary:
Let $(X,\leqslant)$ be a partially ordered set in which every chain has an upper bound, and assume by contradiction that there is no maximal element. Then by the axiom of choice, there exists a function $f$ mapping every chain in $X$ to a strict upper bound. 
For any chain $C\subseteq X$ and $x\in C$, let $P(C,x)$ denote the strict initial segment of $x$ in $C$, that is, $P(C,x)=\{y\in C:y<x\}$.
A subset $A\subseteq X$ is conforming if:


*

*$A$ is well ordered by $\leqslant$,

*For all $x\in A$, $x=f(P(A,x))$
Lemma 1: If $A$ and $B$ are conforming subsets of $X$ and $A \neq B$, then one of these two sets is an initial segment of the other.
Lemma 2:  If $A$ is a conforming subset of $X$ and $x\in A$, then whenever $y < x$, either $y\in A$ or $y$ does not belong to any conforming set.
*Lemma 3: The union $U$ of all conforming subsets of $X$ is conforming.

To reiterate: How can we prove Lemma 3, assuming Lemma 2 and using minimal other set-theoretic results?
 A: We first show that $U$ is a chain. It’s not hard to see that $U$ is a chain. Suppose that $x,y\in U$; there are conforming sets $A_x$ and $A_y$ such that $x\in A_x$ and $y\in A_y$. Lemma $1$ implies that one of these sets is a subset of the other, so without loss of generality suppose that $A_x\subseteq A_y$. Then $x,y\in A_y$, and $A_y$ is a chain, so either $x\le y$, or $y\le x$.
Now we show that $U$ is well-ordered by $\le$. Suppose that $\varnothing\ne S\subseteq U$; we want to show that $S$ has a $\le$-least element. Fix $s\in S$; there is some conforming set $A$ such that $s\in A$. $A$ is well-ordered by $\le$, so let $a=\min\{x\in A\cap S:x\le s\}$; I claim that $a=\min S$. Let $x\in S$; if $x\in A$, then certainly $a\le x$, so suppose that $x\in S\setminus A$. There is some conforming $A_x$ such that $x\in A_x$, and clearly $A_x\nsubseteq A$, so $A$ is an initial segment of $A_x$. Since $x\in A_x\setminus A$, it’s clear that $a<x$ in this case as well and hence that $a=\min S$.
It remains to be shown that if $x\in U$, then $x=f(P(U,x))$. Let $x\in U$; there is a conforming set $A_x$ such that $x\in A_x$. By definition $x=f(P(A_x,x))$, so we’re done if we can show that $f(P(U,x))=f(P(A_x,x))$. It suffices to show that $P(U,x)=P(A_x,x)$. Clearly $P(A_x,x)\subseteq P(U,x)$, so we need only show that $P(U,x)\subseteq P(A_x,x)$, i.e., that if $y\in U$ and $y<x$, then $y\in A_x$. Suppose, then, that $y\in U$ and $y<x$; there is a conforming set $A_y$ such that $y\in A_y$. If $A_y\subseteq A_x$, then certainly $y\in A_x$, and we’re done. Otherwise, $A_x$ is an initial segment of $A_y$, and again $y\in A_x$, since $y<x$.
