# What is the guarentee of existance of the interval of the form $(x_a,a]$ such that $(x_a,a]\cap D=\emptyset$

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?

My effort:-

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 $$(x_a,a]\cap D=\emptyset?$$

We know there is an open set $$U_a$$ containing $$a$$ that misses $$D$$.
Now use that we have the order topology (!): a base for that topology consists of all open intervals $$(x,y)$$ with $$x,y \in X, x < y$$ and also, if $$\min(X)$$ exists, all sets of the form $$[\min(X), x)$$ and if $$\max(X)$$ exists all sets of the form $$(x,\max(X)]$$ as well (otherwise $$\min(X)$$ or $$\max(X)$$ would not be in any basic open set, if we only allowed the open intervals).
So if $$a$$ happens to be $$\max(X)$$ we are done right away, as we have a basic element $$(c,a]$$ inside $$U_a$$.
If not, there must be some $$(c,d)$$ containing $$a$$, that sits inside $$U_a$$ by the definition of a base.
Now we use the well-order: if $$(a,d)$$ is empty, we have no points between $$a$$ and $$d$$ and then $$(c,d)=(c,a]$$ and we are done. If it is non-empty: it has a minimum that we call $$a^+$$ (the smallest element larger than $$a$$) and then $$a \in (c,a^+] \subseteq (c,d) \subseteq U_a$$ and we are done too. (we pick $$x_a=c$$ in all these cases)