This question has been answered in the negative by Gary Gruenhage. I will post a complete answer some time in the future. Here is a sketch of the proof. The existence of an order-preserving map $\psi$ as in the question is equivalent to $\omega_1$ being monotonically orthocompact via open refinements, abbrevaited MO$_o$ (this is Theorem 3.1 in the paper, a link to which is enclosed at the end of this question). What Gary proved is that MO$_o$ implies a certain property called (A$_o$) (defined in terms of certain neignborhoods), and that $\omega_1$ does not have this property (A$_o$). (Update as of August 21, 2020.)
This question has now been published in a journal, see update at the bottom. (This was an older update, 2017 or 2018.)
I posted the following question more than two years ago on MO (and then reposted on MSE), but the answer remains incomplete, so I thought I would rephrase it a bit (to make the statement clearer) and try again.
Let $\omega_1$ be the first uncountable ordinal,
same as the set of all countable ordinals.
$\omega_1=\{\alpha:0\le\alpha<\omega_1\} = \{\alpha:\alpha$ is a countable ordinal$\}$$\omega_1=\{\alpha:0\le\alpha<\omega_1\} = \{\alpha:\alpha {\mathrm{\ is\ a\ countable\ ordinal}}\}$.
Let $\mathcal F$ be the set of all functions
$f:\omega_1\to\omega_1$ that
are:
(a) regressive i.e. $f(\alpha) < \alpha$ for all $0 < \alpha < \omega_1$,
and
(b) non-decreasing (same as $\le$-order-preserving),
i.e.,
if $0\le\alpha \leq \beta<\omega_1$ then $f(\alpha)\leq f(\beta)$ .
Define a partial order $\sqsubseteq$ on $\mathcal F$ by $f \sqsubseteq g$ if
$f(\alpha) \leq g(\alpha)$ for all $\alpha < \omega_1$.
Let $\mathcal K$ be the subset of $\mathcal F$, consisting of functions with
a finite range.
Formally $\mathcal K=\{f\in\mathcal F: |\{f(\alpha):\alpha<\omega_1\}|<\aleph_0\}$.
If I were to make a guess, I would say the answer is no. This question is an order-theoretic restatement of a question from general topology that a co-author and I considered: Whether $\omega_1$ has a monotone interior-preserving open operator $r$, that is, if $\mathcal U$ is any open cover of $\omega_1$, with the order topology, then $r(\mathcal U)$ is an interior-preserving open refinement that covers $\omega_1$, and if $\mathcal U$ refines $\mathcal V$ then $r(\mathcal U)$ refines $r(\mathcal V)$. As usual we would write $\mathcal U\preceq \mathcal V$ if $\mathcal U$ refines $\mathcal V$. In this context $f$ is intended to encode an open cover $\mathcal U(f)=\{0\}\cup\{(f(\alpha),\alpha]:\alpha<\omega_1\}$. Note that if $f\sqsubseteq g$ then $\mathcal U(g)\preceq \mathcal U(f)$.
Update Oct 19, 2018 (and May 21, 2019):
This question has now been published in a journal.
It is Question 3.2 in the following paper:
Serdica Math. J. 44 (2018) (dedicated to the memory
of Professor Stoyan Nedev (1942−2015))
ON MONOTONE ORTHOCOMPACTNESS
S.G. Popvassilev, J.E. Porter
Here is a temporary link from the editors:
http://www.math.bas.bg/serdica/2018/2018-177-186.pdf
(Update as of August 21, 2020.)
This question has been answered in the negative by Gary Gruenhage. I will post a complete answer some time in the future. Here is a sketch of the proof. The existence of an order-preserving map $\psi$ as in the question is equivalent to $\omega_1$ being monotonically orthocompact via open refinements, abbrevaited MO$_o$ (this is Theorem 3.1 in the paper, a link to which is enclosed at the end of this question). What Gary proved is that MO$_o$ implies a certain property called (A$_o$) (defined in terms of certain neignborhoods), and that $\omega_1$ does not have this property (A$_o$).
(Update April 25, 2021.)
I am about to publish an answer here with details of Gary Gruenhage's proof (thus answering the above question is the negative).
