Well-Ordering Theorem implies Hausdorff Maximal Principle

Question

I have a quick question regarding to the proof that well-ordering theorem implies Maximal Principle. In the proof described here https://proofwiki.org/wiki/Well-Ordering_Theorem_implies_Hausdorff_Maximal_Principle The function is defined as

$$\rho(f:S_x \rightarrow \mathcal{P}(X))\begin{cases} f(S_x)\cup\ \{x\} &\mathrm{if}\ P(S_x,x) \\ f(S_x) & \mathrm{otherwise} \end{cases} $$

however, it doesn't seem to make sense to have $$f(S_x)\cup\{x\}$$ as for example if one has $X$ as $\{1, 2, 3, 4, 5, 6\}$ then $$f(1)=\{1\}$$ $$f(2)=\{\{1\},\{2\}\}$$ and so on. Thus $$f(S_3)=\{\{1\},\{\{1\},\{2\}\}\}$$ But I think the proof is intended to mean $$f(S_3)=\{1,2\}$$ So shouldn't the function be defined as the following?

$$\rho(f:S_x \rightarrow \mathcal{P}(X))\begin{cases} \cup_{z \in S_x}f(z)\cup\ \{x\} &\mathrm{if}\ P(S_x,x) \\ \cup_{z \in S_x}f(z) & \mathrm{otherwise} \end{cases} $$

Answer 1

No, $2\neq\{2\}$, so $X\cup\{2\}\neq X\cup\{\{2\}\}$. In other words, $\{1\}\cup\{2\}\neq\{\{1\},\{2\}\}$.

The proof is quite unclear, but here's the idea. Well-order your partial order. This well-order is not necessarily compatible with your partial order, of course. Now recursively go through the well-order, and start collecting elements into a chain: if you reached a certain point, and you've collected a chain so far, add an element if it is comparable to all the things you've collected so far.

At the end of the recursion process we have a chain, and we can prove it is maximal: if we could have added another element to it, why didn't we do it when we got to it on the well-ordering? Well, the only reason is that we couldn't really add it.