In Kunen's Set Theory: An Introduction to Independence Proofs (1980 edition), he defined a Suslin line as the following.
4.1 Definition. A Suslin line is a total ordering, $\langle X,<\rangle$ such that in the order topology, $X$ is c.c.c. but not separable.
Here, the c.c.c. condition means that every collection of pairwise disjoint open subsets of given topological space is countable.
This definition allows some bad cases; for example, $X$ could have gaps in it, or isolated points. Thus the author introduced the following theorem:
4.4 Theorem. If there is a Suslin line, then there is a Suslin line $X$ such that
(1) $X$ is dense in itself $(\textrm{i.e.,} a<b\to(a,b)\neq\varnothing)$, and
(2) no nonempty open subset of $X$ is separable.
Where I feel unclear is the proof of (2); I quote the proof of the book.
Proof. Let $Y$ be any Suslin line. Define an equivalence relation $\sim$ on Y by setting $x\sim y$ iff the interval between them $((x,y)$, $\varnothing$, or $(y,x))$ is separable. Let $X$ be the set of $\sim$-equivalence classes. If $I\in X$, then $I$ is convex; i.e., $x,y\in I\wedge x<y\to(x,y)\subseteq I$. We totally order $X$ by setting $I<J$ iff some (any) element of $I$ is less than some (any) element of $J$.
Note that each $I\in X$ is separable. To see this, let $\mathscr{M}$ be a maximal disjoint collection of nonempty open intervals of the form $(x,y)$ with $x,y\in I$. $\mathscr{M}$ is countable since $Y$ has the c.c.c., so let $\mathscr{M}=\{(x_n,y_n):n\in\omega\}$. Since $x_n\sim y_n$, let $D_n$ be a countable dense subset of $(x_n,y_n)$. Let $D=\bigcup_nD_n$; then $D$ is dense in $\bigcup_n(x_n,y_n)$. If $z\in I$ and $z\in(x,y)\subseteq I$, then $(x,y)$ intersects some $(x_n,y_n)$ by maximality of $\mathscr{M}$; this implies $z\in\overline{D}$ unless $z$ is the first or last element of $I$. Thus, $D$ together with the first and last elements of $I$ (if $I$ has a first or last element) forms a countable dense subset of $I$.
To see that $X$ is dense itself, suppose $I<J$ but $(I,J)=\varnothing$. Pick $x\in I$ and $y\in J$, then $(x,y)\subseteq I\cup J$, which is separable, so $x\sim y$, a contradiction.
To verify (2), it is sufficient to see that $(I,J)$ is not separable whenever $I<J$. Suppose it were. Let $\{K_n:2\le n<\omega\}$ be dense in $(I,J)$, and let $K_0=I$, $K_1=J$.
In $Y$, let $D_n$ be a countable dense subset of $K_n$, then $\bigcup_nD_n$ is dense in $\bigcup\{L:I\le L\le J\}$, so points of $I$ are equivalent to points of $J$, a contradiction. (the rest omitted)
I'm not sure that, in the last part of the proof, $\bigcup_nD_n$ is dense in $\bigcup\{L:I\le L\le J\}=:N\subseteq Y$.
If we take distinct two points $x,y\in N$ with $x<y$ which belong to the same $\sim$-equivalence class that does not equal to any $K_n$ - of course, I don't know whether $(x,y)$ is empty or such equivalence class exists - it seems to be that there is no element of $\bigcup_nD_n$ between $x$ and $y$. Is there any mistake in my opinion, or some explanation for this theorem? I appreciate in advance.
Indeed, Kunen's argument is incomplete in the way that you describe. To fix it, you need to use the fact that $Y$ is ccc. This implies that all but countably many elements of $X$ must have at most two elements: any element of $X$ that contains three or more elements contains a (nonempty) open interval of $Y$, and so if there were uncountably many such elements of $X$ you would get an uncountable family of disjoint open intervals of $Y$, which is impossible since $Y$ is ccc.
So, in the proof of (2), actually only countably many elements of the interval $(I,J)$ have more than two elements, and you can choose $(K_n)$ to include all of those countably many elements. Then the problem you describe cannot arise (such an interval $(x,y)$ would always be empty), and it is easy to verify that $\bigcup D_n$ is indeed dense.