I am trying to understand a proposition which goes like this:
Let $(X,\precsim)$ be a nonempty totally preordered set.
Let $(\hat{X}, \precsim)$ be the set of all equivalence classes in $(X,\precsim)$, i.e. $X$'s quotient set endowed with the same order as $X$ (thus clearly a totally-ordered set).
Then $\exists f: \hat{X} \rightarrow \mathbb{R}$, where $f$ is order-embedding, iff $\exists X^* \subset \hat{X}$ such that $X^*$ is at most countable and order-dense (close-packed) in $\hat{X}$. ($\mathbb{R}$ is ordered as usual).
I just can't understand how to prove this nor get any intuition why this is so. (The fact that it starts with a proset and not a toset is irrelevant to my lack of understanding, I just wanted to state it in full.)
Edit: Order-embedding, not order-isomorphic. Sorry for the confusion. (bis)
First let me restate the proposition (as I understand it) in human language, rather than try to imitate your notation.
Proposition. Let $X$ be a totally ordered set. The following statements are equivalent:
(1) there is a set $D\subseteq X$ which is (at most) countable, and which is dense in $X,$ in the sense that $D\cap[a,b]\ne\emptyset$ whenever $a,b\in X,\ a\lt b.$
(2) $X$ is order-isomorphic to a subset of $\mathbb R.$
$\underline{\text{(1)}\implies\text{(2)}}:$ Let $D=\{d_n:n\in\mathbb N\}.$ Define $f:X\to\mathbb R$ by setting $$f(x)=\sum_{d_n\lt x}2^{-n}+\sum_{d_n\le x}2^{-n}.$$ It is easy to see that $x,y\in X,\ x\lt y\implies f(x)\lt f(y).$
$\underline{\text{(2)}\implies\text{(1)}}:$ (Axiom of choice needed here.) Without loss of generality, we assume that $X\subseteq\mathbb R.$ For each rational number $r,$ let $x_r$ be the greatest element of $X\cap(-\infty,r)$ if such exists, and let $y_r$ be the least element of $X\cap(r,\infty)$ if such exists. For each rational interval $(r,s)$ such that $X\cap(r,s)\ne\emptyset,$ choose an element $z_{r,s}\in X\cap(r,s).$ Let $D$ be the set of elements $x_r,y_r,z_{r,s}$ so chosen. Clearly $D$ is a countable dense subset of $X.$