This problem occurred to me when I was solving problem 112 in A. Shen and N. K. Vereshchagin, Basic Set Theory (AMS 2002)), just as in this post Can this proof of existence of a Hamel basis using transfinite recursion be shortened/simplified? . So I will quote some parts from that post here.
Let $V$ be vector space with some partial order $\leq$ over it that is isomorphic to $\mathbb{R}$ with its standard ordering $\leq$.
Let be $f: V \to \mathscr{P}(V)$ such function that $$ f(x) = \begin{cases} \bigcup_{w < x} f(w) & \text{if } x \in \operatorname{span}\left(\bigcup_{w < x} f(w)\right) \\ \bigcup_{w < x} f(w) \cup \{x\} & \text{if } x \notin \operatorname{span}\left(\bigcup_{w < x} f(w)\right). \end{cases} $$
We could use the theorem from the book to prove that such function does exist if $\leq$ was well-order over $\mathbb{R}$:
Let $A$ be a well-ordered set, and $B$ an arbitrary set. Let a recursive rule be given, that is, a mapping $F$ whose arguments are an element $x \in A$ and a function $g: [0, x) \to B$, and whose value is an element of $B$. Then there exists exactly one function $f: A \to B$ such that $$ f(x) = F\big(x, f|_{[0, x)}\big) $$ for all $x \in A.$
But of course $\leq$ is not well-order over $\mathbb{R}$. So I think such function doesn't exist (in general case or maybe always, I'm not sure), but how to see and prove it?