Understood How Zorn's Lemma Implies the Axiom of Choice

Many math books that I’ve read referred me to other books for the proof of the equivalence of the Axiom of Choice and Zorn’s Lemma. This afternoon, I spent more than two hours to understand the proof that the later implies and former in Topology written by Davis.

1. Set up a non-empty partially ordered set $(\mathcal{P},\le)$
2. Let $\mathcal{T}$ be any non-empty chain in $\mathcal{P}$.
3. Prove that $\cup \mathcal{T} \in \mathcal{P}$.
4. Apply Zorn’s Lemma to get a maximal element $g \in \mathcal{P}$.
5. Use the maximality of $g$ to claim that the domain of $g$ equals the family of non-empty subsets $(S_i)_{i \in I}$ from which elements $(x_i)_{i \in I}$ are chosen.

In the book, $\le$ means function extension, and

$\mathcal{P} = \{f \mid f \text{ is a function, } dom\,f \subseteq (S_i)_{i \in I}, f(x) \in x \,\forall x \in dom\,f\}.$

Step (3) is proved step-by-step according to the definition of $\mathcal{P}$. Usually, suppose that $(x_1,y_1),(x_2,y_2) \in f$, $x_1 = x_2 \implies y_1 = y_2$. The book uses the contrapositive form of this statement. I was stuck at the sentence Now, for a set to be an element of $dom\cup\mathcal{T}$, it must be an element of some member of $\mathcal{T}$. At first, I omitted the phrase "some member of", and stopped for half an hour. Reading the next sentence Hence $dom\cup\mathcal{T} \subseteq (S_i)_{i \in I}$, I knew how to interpret the sentence where I was stuck: if $S \in dom\cup\mathcal{T}$, $\exists f \in \mathcal{T}$ such that $S \in dom\,f$. Since $f \in \mathcal{T} \subseteq \mathcal{P}$, $dom\,f \subseteq (S_i)_{i \in I}$. Then $S \in (S_i)_{i \in I}$, and thus $dom\cup\mathcal{T} \subseteq (S_i)_{i \in I}$.