My questions from the same excerpt quoted here,
Proof. $\ \ $ Recall that a subspace $Y$ of $L$ is said to be convex if for every pair of points $a,b$ of $Y$ with $a<b$, the entire interval $[a,b]$ of points of $L$ lies in $Y$. We prove that if $Y$ is a convex subspace of $L$, then $Y$ is connected.
$\quad$ So suppose that $Y$ is the union of the disjoint nonempty sets $A$ and $B$, each of which is open in $Y$. Choose $a\in A$ and $b\in B$: suppose for convenience that $a<b$. The interval $[a,b]$ of points of $L$ is contained in $Y$. Hence $[a,b]$ is the union of the disjoint sets $$A_0=A\cap[a,b]\quad{\rm and}\quad B_0=B\cap[a,b],$$ each of which is open in $[a,b]$ in the subspace topology, which is the same as the order topology. The sets $A_0$ and $B_0$ are nonempty because $a \in A_0$ and $b\in B_0$. Thus, $A_0$ and $B_0$ constitute a separation of $[a,b]$.
$\quad$ Let $c=\sup A_0$. We show that $c$ belongs neither to $A_0$ nor to $B_0$, which contradicts the fact that $[a,b]$ is the union of $A_0$ and $B_0$.
$\quad$ Case 1. Suppose that $c\in B_0$. Then $c\neq a$, so either $c=b$ or $a<c<b$. In either case, it follows from the fact that $B_0$ is open in $[a,b]$ that there is some interval of the form $(d,c]$ contained in $B_0$. If $c=b$, we have a contradiction at once, for $d$ is a smaller upper bound on $A_0$ than $c$. If $c<b$, we note that $(c,b]$ does not intersect $A_0$ (because $c$ is an upper bound on $A_0$). Then $$(d,b]=(d,c]\cup (c,b]$$
My doubt is, if we have that $(d,c]$ is contained purely in $B_o$, then wouldn't it mean we have a contradiction in the beginning itself since $d$ can not be in $A$ but is greater than everything in it?
I understand already why the set should exist from here but I don't understand why the case work is done.
