Countable order-dense subset in a linear order I ended up posting an answer to my own question, and I believe I knew this answer before I posted this question, but forgot about it.  At any rate, here it is for the record. 
My question is inspired by the following one by @lu_sin. Is there a countable order-dense family of open subsets of $(0,1)$?
While I answered the above question (and my answer was accepted), the question remains open under a slightly different definition of order-dense, as indicated in the comments over there. While trying to answer the modified version of the question (which I believe ought to have something to do with $(0,1)$ being hereditarily Lindelöf and hereditarily separable), I came up with the following question. 
Let $(X,<)$ be a linear (same as total) order. Assume that: 
$(*)\ \ \ $ $X$ has no uncountable well-ordered subset (under the order induced by $<$), and no uncountable reverse well-ordered subset. 
Call a set $D\subseteq X$ order-dense in $X$ if whenever $\{x,y\}\subseteq X$ with $x<y$ then there is $z\in D$ such that $x\le z\le y$. (Note that under this definition the integers are order-dense in themselves, which would not be the case if we required that $x<z<y$. In that other question I find the version with $\le$ more difficult to answer, which is the reason I use $\le$ here.) 
Question. Assuming condition $(*)$ as above, is there a countable set $C$ which is order-dense in $X$? 
I could answer it under a certain "local" assumption, as explained below (but I think this assumption makes the answer rather easy, and I do not know the answer in general). 
(On third reading I realized that I have a gap in the construction below, but I will leave  it as is for now, perhaps it illustrates some of the ideas I was thinking about.) 
Call an $x\in X$ $r$-good (for right-good) if either (a) $x=\max X$ (if it exists), or (b) there is $x^+>x$ and a countable set $D=D(x,x^+)$ that is order-dense in $[x,x^+]$. Analogously, call an $x\in X$ $l$-good if either (a) $x=\min X$ (if it exists), or (b) there is $x^-<x$ and a countable set $D=D(x^-,x)$ that is order-dense in $[x^-,x]$. If each $x$ in $X$ is good (i.e. both $r$-good and $l$-good) then the answer to the above question is yes. 
Indeed starting with any $x_0\in X$ we could get an increasing countable transfinite sequence 
$S=\{x_\alpha:\alpha<\gamma\}$ for some $0<\gamma<\omega_1$ such that $\sup S=\sup X$, and for each $\alpha$ there is a countable $D_\alpha$ order-dense in $[x_\alpha,x_{\alpha+1}]$ (except only if $x_\alpha=\max X$). Similarly, there 
is a decreasing countable transfinite sequence $T=\{y_\alpha:\alpha<\delta\}$ for some $0<\delta<\omega_1$ such that $\inf T=\inf X$, $y_0=x_0$, and for each $\alpha<\delta$ there is a countable $C_\alpha$ order-dense in $[y_{\alpha+1},y_\alpha]$ (unless $y_\alpha=\min X$). Then we could let $C$ to be the union of all $D_\alpha,\alpha<\gamma$, and 
all $C_\alpha,\alpha<\delta$ (also throw in $S$ and $T$ and the endpoints of $X$, if any). 
There is a gap in the above argument, I was implicitly assuming that if $\alpha>0$ is limit then $x_\alpha=\sup_{\beta<\alpha}x_\beta$ 
(and similarly $y_\alpha=\inf_{\beta<\alpha}y_\beta$.) If
$\alpha>0$ is limit and $\sup_{\beta<\alpha}x_\beta$ is a gap of $X$ (i.e. $\sup_{\beta<\alpha}x_\beta$ does not exist in $X$ because $X$ is not complete) then I do not know how I want to define $x_\alpha$. 
The answer is likely to be known, I would be grateful for a reference too. 
Extra comments on condition $(*)$. 
For the record, condition $(*)$ was first stated as follows. 
For every $Y\subseteq X$ there is a countable (finite allowed) $Z\subseteq Y$ such that $\sup Y=\sup Z$ and there is a countable $W\subseteq Y$ such that $\inf Y=\inf W$. (Note that we may replace $Z$ with $Z\cup W$ and require, equivalently, that for every $Y\subseteq X$ there is a countable $Z\subseteq Y$ such that $\sup Y=\sup Z$ and $\inf Y=\inf Z$.) It seems we could drop the requirement that $Z\subseteq Y$ and $W\subseteq Y$. Just require that for every $Y\subseteq X$ there is a countable (finite allowed) $Z$ such that $\sup Y=\sup Z$ and there is a countable $W$ such that $\inf Y=\inf W$. Then I edited it as I realized: 
I may have made things more complicated than necessary, will think, perhaps I should have just said that $X$ has no subset that is order-isomorphic to $\omega_1$, or reverse order-isomorphic to $\omega_1$. 
One example of a set that satisfies condition $(*)$ is the open unit interval $(0,1)$ with the usual order $\le$. 
Another example (related to the question that was linked): If $X=\mathcal V$ where $\mathcal V$ is a nested collection of open subsets of $(0,1)$ (i.e. a linear order under set-inclusion), then $\mathcal V$ satisfies condition $(*)$, which could be shown (I believe) using that $(0,1)$ is hereditarily Lindelöf and hereditarily separable. 
I realize I could have stated my question in yet another different, and apparently more general way (closer to the statement of the linked question). Suppose $P$ is a partial order and $X$ is a subset which is a chain, i.e. a linear order (under the induced order from $P$). Call a set $D\subseteq P$ dense in $X$ (where we do not require that $D\subseteq X$) if whenever $\{x,y\}\subseteq X$ with $x<y$ then there is $z\in D$ such that $x\le z\le y$. Question. If $X$ satisfies condition 
$(*)$ above, is there a countable set $C\subseteq P$ which is order-dense in $X$ (where we do not require that $C\subseteq X$)?
I tend to believe that the two versions of the question are equivalent, but did not attempt to verify all details. 
Here is a related example, let $\mathcal V=\{(-x,x):0<x<1\}$, a nested collection of open sets in $(-1,1)$. Let $D=\{(-r,r):0<r<1, \ r\in\Bbb Q\}$, where $\Bbb Q$ denotes the rational numbers. Then $D$ is countable and order-dense in $\mathcal V$ (under either definition). 
This question is such a mess ... I feel like I have not done my homework, but I will leave it as is for now, perhaps come back to it later. 
 A: Rather than hassle the problem per se, consider simply the upper portion.
Let S be a complete linear order.
Assume for all A subset S,
exists countable B subset A with sup A = sup B.
Is there a countable, weakly order dense subset of S?  
Is the existence of such a set equivalent to
for all A subset S, A is not order isomorphic to $\omega_1$?  
D is weakly order dense, when
for all a,b in S, (a < b implies exists x in D with a <= x <= b).  
Assume S is a complete linear order let A be order isomorphic to $\omega_1$.
Then A subset S does not have the property
exists countable subset B, sup A = sup B.
Thus, the nonexistence of such a set is not equivalent to a weakly order dense subset of S + the assumptions for S, but a result of those assumptions.  
A: Hmm, I think the answer is no, and is rather easy (in the generality in which I asked the question), and I knew it when I tried to answer that other linked question but forgot about it :). I do not know what I want to do of this question, feels like there ought to be a better version of the question where the answer might be interesting. 
So here is my attempted answer. 
Let $X=\{(-x,x):0<x<1\}\cup\{[-x,x]:0<x<1\}$. Notice this is a linear order under set-inclusion. Condition $(*)$ is easy to verify (boils down to using that $(0,1)$ with $\le$ has no uncountable well-ordered, or uncountable reverse well-ordered subset). 
Yet, if $D\subseteq X$ is a set which is order-dense in $X$ (and which apparently I should call weakly order-dense, as in the answer by @WilliamElliot) then, for each $x$, either $(-x,x)$ or $[-x,x]$ (or both) must be in $D$, showing that $D$ must necessarily have the cardinality of the continuum. 
Copied/edited from @WilliamElliot's answer:
Definition. $D$ is weakly order-dense in $X$ if for all $x,y$ in $X$ with $x<y$ there is $z\in D$ such that $x\le z\le y$. 
This just makes the linked question of @lu_sin more interesting to me (in its version about weakly order-dense, i.e when, in that question, we only require $V \subseteq U \subseteq V'$ instead of $V \subsetneq U \subsetneq V'$). The linear order there consists of open sets only (open subsets of $(0,1)$ ordered by set inclusion). Is there a countable order-dense family of open subsets of $(0,1)$?
