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$\}$.
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 $\mathcal K=\{f\in\mathcal F: |\{f(\alpha):\alpha<\omega_1\}|<\aleph_0\}$.

Question:
Is there a $\sqsubseteq$-order-preserving map (same as a $\sqsubseteq$-non-decreasing map)
$\psi : \mathcal F \to \mathcal K$, i.e if $f \sqsubseteq g$ then $\psi(f) \sqsubseteq \psi(g)$, and with
the additional property that $\psi(f) \sqsupseteq f$ for all $f\in \mathcal F$ ?

Let me summarize some comments made at MO, clarifying certain partial answers.

Partial answer (A). Since every $f\in\mathcal F$ is regressive and non-decreasing, it must be eventually constant and reach its maximal value $\mu_f=\max\{f(\alpha):\alpha < \omega_1\}$. One is tempted to define $\psi(f)(\alpha)=\mu_f$ for all $\alpha$. The problem is that this is not regressive: We have $\psi(f)(\alpha)<\alpha$ only when $\alpha>\mu_f$, but I insist that $\psi(f)(\alpha)<\alpha$ whenever $0<\alpha<\omega_1$.

Partial answer (B). If we drop the requirement that $\psi$ be a $\sqsubseteq$-non-decreasing map then the answer by @NoahS below works, as well as one of my comments below, which I move here. As above let $\mu_f=\max\{f(\alpha):\alpha < \omega_1\}$ and let $\gamma_f=\min\{\alpha:f(\alpha)=\mu_f\}$. (Then $f(\alpha)=\mu_f$ for $\alpha\ge\gamma_f$, and $f(\alpha)<\mu_f$ for $\alpha<\gamma_f$. Usually $\mu_f<\gamma_f$ unless $\mu_f=0=\gamma_f$.) Let $\alpha_{0,f}=\mu_f$. If $\mu_f\ge1$ then let $\alpha_{1,f}=f(\alpha_{0,f})<\alpha_{0,f}$. There is a non-negative integer $n_f$ such that $\alpha_{k+1,f}=f(\alpha_{k,f})<\alpha_{k,f}$ for $k<n_f$, and $\alpha_{n_f,f}=0$. Define $\psi(f)$ as follows. If $\alpha>\alpha_{0,f}$ then let $\psi(f)(\alpha)=\alpha_{0,f}=\mu_f$. If $\alpha_{k+1,f}<\alpha\le\alpha_{k,f}$ then let $\psi(f)(\alpha)=\alpha_{k+1,f}$. (Formally also $\psi(f)(0)=0$, but in general each function in $\mathcal F$ being regressive must take value $0$ at $1$, and being non-decreasing must take value $0$ at $0$ as well.) Then $\psi(f)\in\mathcal K$ and $\psi(f)\sqsupseteq f$.

So partial answer (A) above achieves that $\psi(f)$ has a finite range, and $\psi(f) \sqsubseteq \psi(g)$ whenever $f \sqsubseteq g$, and also $\psi(f) \sqsupseteq f$. It almost achieves that $\psi(f)$ is regressive, but not quite, and it follows that $\psi(f)$ is not in $\mathcal K$ unless $\mu_f=0$. (One could perhaps say that $\psi(f)$ is "regressive on a tail" only, which might in a different context be good enough, but the requirement in my question is that $\psi(f)(\alpha)<\alpha$ whenever $0<\alpha<\omega_1$.) On the other hand, partial answer $B$ achieves that $\psi(f)\in\mathcal K$ (in particular both that $\psi(f)$ is regressive and has a finite range), and $\psi(f) \sqsupseteq f$ for all $f\in \mathcal F$, but not necessarily that $\psi(f) \sqsubseteq \psi(g)$ whenever $f\sqsubseteq g$. It is not clear to me if we could achieve all conditions simultaneously. Edit. Following a comment, let me clarify why in partial answer $B$ we need not have $\psi(f) \sqsubseteq \psi(g)$ whenever $f\sqsubseteq g$. Fix any ordinals $0<\beta<\delta<\nu<\omega_1$. Let $f(\alpha)=g(\alpha)=0$ if $0\le\alpha<\nu$. Let $f(\alpha)=\beta$ and $g(\alpha)=\delta$ if $\alpha\ge\nu$. Clearly $f\sqsubseteq g$. Then $\psi(f)(\alpha)=\beta$ if $\alpha>\beta$, and $\psi(f)(\alpha)=0$ if $0\le\alpha\le\beta$ (where $\psi$ is as in partial answer $B$). While $\psi(g)(\alpha)=\delta$ if $\alpha>\delta$, and $\psi(g)(\alpha)=0$ if $0\le\alpha\le\delta$. In particular, if $\beta<\alpha\le\delta$ then $\psi(g)(\alpha)=0<\beta=\psi(f)(\alpha)$, so $\psi(f)\not\sqsubseteq \psi(g)$.

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 $\mathcal U(g)\preceq \mathcal U(f)$.

Thank you!

Original version of this post:

Let $\omega_1$ be the first uncountable ordinal, same as the set of all countable ordinals. Let $F$ be the set of all functions $f$ from $\omega_1$ minus singleton $0$ into $\omega_1$ that are (a) regressive i.e. $f(\alpha) < \alpha$ for $0 < \alpha < \omega_1$, and (b) order-preserving (same as non-decreasing) i.e. $f(\alpha)\leq f(\beta)$ if $\alpha \leq \beta$. Define a partial order on $F$ by $f \leq g$ if $f(\alpha) \leq g(\alpha)$ for all $0<\alpha < \omega_1$. Let $K$ be the subset of $F$, of functions with a finite range.

Question: Is there an order-preserving homomorphism $h : F \to K$, i.e if $f \le g$ then $h(f) \le h(g)$, and with the additional property that $f \le h(f)$ ?

(I had dropped (b) in my first post, but comments below show that it is essential. Also, I did mean the functions $f$ must be regressive, when I had imprecisely said decreasing, in the original statement. I added the gn tag since the question stated is an order-theoretic translation of a question from general topology: Whether $\omega_1$ has a monotone interior-preserving open operator.)

Edit April 28, 2014: The answer by Noah S below is correct but incomplete (for each $f\in F$ it finds $h(f)\in K$ with $f \le h(f)$ but does not consider whether $h(f) \le h(g)$ when $f\le g \in F$). The question is open. Thank you

Edit April 12, 2015. I reposted at MSE

Update Oct 19, 2018 (and May 21, 2019):
This questions has been included 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