Proving Hausdorff maximality principle without Choice

I have heard that it is possible to prove a variant of the Hausdorff Maximality Principle without the axiom of choice. This is called "Hausdorff Maximality Principle for well-ordered partial orders" and it says that for every partial order (P, ≤) ∈ V with the property that there exists a well- order ≺ on the underlying set P, there is an inclusion-maximal chain X in (P, ≤). How could I prove it using only ZF and not Choice?

• Do you happen to have heard that in problem set 5? – Alessandro Codenotti Nov 14 '18 at 9:18

Recursively construct $$(C_\xi \mid \xi \in \mathrm{Ord})$$ as follows:
Let $$C_0 := \emptyset$$ and given $$C_\xi$$ either $$C_\xi$$ is a $$\subseteq$$-maximal chain in $$(P; \le)$$ in which case we stop the construction or otherwise $$C_{\xi +1} := C_\xi \cup \min_{\prec} \{ p \in P \setminus C_\xi \mid \forall c \in C_\xi \colon p \le c \vee c \le p \}.$$ For limit $$\lambda \in \mathrm{Ord}$$, we let $$C_{\lambda} := \bigcup_{\xi < \lambda} C_{\xi}$$.
It is easy to verify that each $$C_\xi$$, if defined, is a chain through $$(P; \le)$$ and that there is some $$\xi < H(P)$$ such that $$C_{\xi}$$ is a maximal chain.
(Here $$H(P)$$ is the least ordinal $$\alpha$$ such that there is no injection $$i \colon \alpha \to P$$. Since $$P$$ has a well-ordered, we have that $$H(P) = \mathrm{card}(P)^+$$ but for the sake of this proof it's actually more natural to think about it as $$H(P)$$ -- the Hartogs number of $$P$$.)