Let $X$ be a totally ordered set. I say that a point $x$ is "left-isolated" if there is some $\alpha < x$ such that there is only one point $y$ which satisfies $\alpha < y \leq x$, that is, $y=x$. I assume that there is some countable subset $D \subset X$ such that any non-empty open interval $(\alpha,\beta)=\{x \in X ; \alpha < x < \beta\}$ meets $D$ (so that $D$ is dense if $X$ is endowed with the topology generated by open intervals). It is obvious that the set of "isolated points" (that is, points $x$ such that for some $\alpha,\beta$ we have $(\alpha,\beta)=\{x\}$) is countable.
Question: is the set of left-isolated points countable also?
Not necessarily. Let $\Bbb P=\Bbb R\setminus\Bbb Q$ be the set of irrational numbers, and let
$$\begin{align*} X&=\{\langle x,k\rangle\in\Bbb R\times\{0,1\}:x\in\Bbb P\text{ or }k=0\}\\ &=\big(\Bbb R\times\{0\}\big)\cup\big(\Bbb P\times\{1\}\big)\;, \end{align*}$$
and let $\preceq$ be the restriction to $X$ of the lexicographic order on $\Bbb R\times\{0,1\}$. Then $\Bbb Q\times\{0\}$ is a countable dense subset of $X$ in the order topology induced by $\preceq$, but each $\langle x,1\rangle$ with $x\in\Bbb P$ is left isolated: its left neighbor is $\langle x,0\rangle$.
Informally, I’ve just split each irrational into two adjacent points.