In the Garling's proof of 'AC implies Zorn's lemma'

$ \newcommand{\powset}{\mathcal{P}} \newcommand{\emtset}{\varnothing} $ (The post became longer than I expected... but help me) Garling showed 'AC $ \Rightarrow$ Zorn's lemma' in Theorem 1.9.1 in his book A Course in Mathematical Analysis Volume 1: Foundations and Elementary Real Analysis (2013).

Assume AC. For any nonempty poset $(X, \le)$, if every nonempty chain of $X$ has an upper bound, then there is a maximal element in $X$.

Notations: For a subset $A$ of $X$, $A'$ denotes the set of strict upper bounds of $A$. A subset $A$ of a totally ordered set $(S, \le)$ is an initial segment of $S$ if for all $x \in S$ and all $y \in A$, $x\le y$ implies $x \in A$.

A choice function $c: \powset(X)\setminus \{\emtset\} \to X$ exists such that for all nonempty subset $A$ of $X$, $c(A) \in A$ by AC.

He defined an interesting set:

$T$ is the set of chains $C$ in $X$ such that if $D$ is an initial segment of $C$, and $D \neq C$, then $c(D')$ is the least element of $C\setminus D$.

The definition of $T$ is clear but the proofs using it make me really confused. Following his proof, I could't really understand some of his statements appearing in the proposition:

Lemma A.1.6 Suppose that $C, D \in T$, and that $C$ is not contained in $D$. Then, $D$ is an initial segment of $C$.

To show this, he defined the set $$ E = \{ x \in C \cap D ~|~ \text{if $y \in x$, then $y \in C$ iff $y \in D$}\} $$ I could easily show that $E$ is an initial segment of both $C$ and $D$. Moreover, because $E \subseteq D$ and $\neg C \subseteq D$, we have $E \neq C$. Therefore, because $C \in T$, $c(E')$ is the least element of $C\setminus E$. There's no problem so far. The confusing point is the following:

Suppose $E \neq D$. Then, $c(E')$ is the least element of $D\setminus E$. But this implies that $c(E') \in E$, giving a contradiction.

How on earth could $c(E')$ be an element of $E$? I can't find any clue of that conclusion. First, $E'$ and $E$ are disjoint and $c(E') \in E'$ by definition. The condition that $c(E')$ is the least element of $C \setminus E$ also does not imply that $c(E') \in E$.

I've been holding this for whole two days. I am sticking to his proof because it seems to use only elementary mathematical notions I can understand for now.

You have $E = \{ x \in C \cap D ~|~ \text{if $y \le x$, then $y \in C$ iff $y \in D$}\}$

You have shown that $E$ is an initial segment of $C\cap D$ and that $c(E')$ is least in $C\setminus E.$ If $E\neq D,$ then $c(E')$ is also least in $D\setminus E.$

Of course, then $c(E')\in C\cap D.$

Now, if $y= c(E')$ then since $c(E')\in C\cap D,$ clearly $y \in C$ iff $y \in D.$

And if $y<c(E')$ then $y\in E$ and again, since $E$ is an initial segment of $C\cap D,\ y \in C$ iff $y \in D.$

Thus, $c(E')\in E.$