TITLE: Distinct well-orderings of the same set
QUESTION [36 upvotes]: An easy consequence of the Erdős-Dushnik-Miller theorem $\kappa\to(\kappa,\omega)^2$ is the following, that  will denote $(*)$ (it appears as an exercise in Kunen's book, it was probably mentioned explicitly earlier by Dushnik-Miller or Erdős, but I haven't found a reference):

If $X$ is infinite and $<_1$ and $<_2$
  are two well-orderings of $X$, then
  there is a subset $Y$ of $X$ with
  $|Y|=|X|$ where $<_1$ and $<_2$
  coincide. 

The proof uses choice, but it follows that it holds in ZF: We may as well assume $X$ is an ordinal $\kappa$, so we can code $<_1$ and $<_2$ with a set $A\subseteq \kappa^2$, and argue in $L[A]$, which is a model of choice.
This proof has always looked bizarre to me. I remember a few years ago discussing this with Clinton Conley, and I believe we found a combinatorial argument, but I don't recall how it went, so I am not sure. 

What I am asking then is for a
  combinatorial proof of $(*)$ in ZF.

(``Combinatorial'' may signify several things, what I mostly mean is that no metamathematics appear in the proof.)
Perhaps there is some kind of inductive argument (I vaguely recall that's what we had), but I don't think it can be completely straightforward. For example, $X=\omega_1$ could have cofinality $\omega$, so the obvious approach may run out of room before $\omega_1$ steps.

REPLY [4 votes]: Just a little update on this question.  The argument in Andres' answer actually gives the following (in principle stronger but who knows) fact in ZF: any ordinal-valued function on a wellordered set $X$ is nondecreasing on a subset equinumerous with $X$.  This doesn't seem particularly revolutionary.  I don't see an adaptation of the argument that gives the full E-D-M theorem in the choiceless setting.