Proof of Zorn's lemma clarification

Let $U$ be the union of all conforming sets.

We need to show that $\le$ is a well-order on $U$. So let $V\subset U$ be a non-empty subset and $v\in V$. There exists a conforming set $A$ with $v\in A$. As $A$ is well-ordered, let $a=\min(A\cap V)$. Then $a=\min V$ because for any $b\in V$, we find a conforming $B$ with $b\in B$. As one of $A,B$ is an initial segment of the other, $A\cup B$ is comforming, hence by its well-order either $b\ge a$ or $b\in A$ and $b\in V$, contradicting $a=\min(A\cap V)$.

We also need to show the second property. So let $x\in U$. Then $x\in A$ for some conforming $A$. Then $P(A,x)=P(U,x)$ because $u\in U$ and $u<x$ implies $u\in A$. It follows that $x=f(P(U,x))$.

Now as $U$ is conforming, every conforming $A$ is either $=U$ or an initial segment of $U$.

The premise can be equivalently rephrased as $$ P_1 :\Leftrightarrow \forall A, B \subseteq X: [\neg A \subseteq B \Rightarrow \exists b \in A: P(A,b) = B] $$

We have to show that 1) for any nonempty subset $S$ of $U$, $S$ has a least element; and that 2) $$ \forall x \in U: x = f(P(U,x)) $$

I used element-chasing for the proof and I didn't use the property 'We observe that ... any conforming set.' The proposition 'The union $U$ of all ...' can be shown directly from $P_1$.

I will list some useful properties for the proof of the statement '$U$ are well-ordered by $\le$':

1. For any $x \in U$, there is a conforming subset of $X$ that contains $x$.
2. For any conforming sets $A$ and $B$, if there is an element $x \in A$ that is not in $B$, it follows that $\neg A \subseteq B$; thus, $\exists b \in A: P(A,b) = B$;
3. For any subset $S$ of $X$ and any conforming set $A$, the intersection $A \cap S$ has a least element because $A \cap S \subseteq A$ and $A$ is well-ordered; $\exists x \in A \cap S: \forall y \in A \cap S: x \le y $;
4. If $x$ and $y$ are in a conforming set, they are comparable, that is, $\neg x < y$ implies $y \le x$ because conforming sets are totally ordered;
5. For any nonempty set $A$, if it is given that $x \in A$ and $ x \notin A \lor Q $, we have $Q$. (resolution) This is used for the case $y \notin P(A,b)$, i.e., $y \notin A \lor \neg y < b$.

The proof is just repeated applications of the above properties. Start from a subset $S$ of $U$.

You can find the proof of the part involving the choice function $f$ from Brian M. Scott.