Let $(X,\leq)$ be a well-ordered set, Let $\mathscr T$ be the order topology on $X$. Then $(X,\mathscr T)$ is normal.
Proof. We know that $(X,\mathscr T)$ is Hausdorff. Then, It is $T_1$. Every interval in $X$ is of the form $(a,b]$ is open in $X$.
Since $(X,\leq)$ is well-ordered set, $X$ has a least member $a_0$. Suppose $C$ and $D$ are disjoint closed subsets of $X$.
Our aim is to find two disjoint open subsets of $X$, $U$ and $V$.
How does the underlined statement be true?
Since $D$ is closed, $X\setminus D$ is open. We also know that $C\cap D=\emptyset$. So, $C\subset X\setminus D$. I know that, $\forall a\in C, a\ge a_0.$ Also there is an open set $U_a$ containing $a$ which is a subset of $X\setminus D.$ I don't know anything other than this. How do I prove there is an interval $(x_a,a]$ such that $(x_a,a]\cap D=\emptyset?$