Is this topological space normal? Let $(X,\leq)$ a totally ordered set with, at least two elements. Let $\mathcal B = \{B_x\mid x\in X\}$ with $B_x=\{y\in X\mid x\leq y\}$ and $T$ the topology in $X$ generates by $\mathcal B$. I have to study in $(X,T)$ the separation axioms, the compacity, conexion, local conexion and conexion by paths.
My attempt
I have proof that $\mathcal B$ generate $T$ as a basis and that the space is not  $T_1$, but it is $T_0$. Also, that the space is compact iff $X$ has a minimum element and that it is connected (because there not exists non-empty disjoints open sets), locally connected and connected by paths.
But, I have troubles to prove if the space is normal or not. If there exists two non-empty disjoints closed sets, I know that the space is not normal by the same argument because it is connected. And, by the other way, if there not exists two non-empty disjoint closed sets, the space would be normal trivialy by definition.
If you could guide me, I appreciate that.
 A: $\let\subset\subseteq$The existence of nonempty disjoint closed sets is equivalent to the existence of proper open subsets whose union is $X$, i.e., the existence of open $U, V \subsetneq X$ with $U \cup V = X$.
Let us show that this is not possible. We prove a stronger claim.
Claim. Let $U, V \subset X$ be open. Then, either $U \subset V$ or $V \subset U$.
Proof. Suppose $U \not\subset V$. We show $V \subset U$.
Pick $u \in U \setminus V$. Since $U$ is open, there is a basis element $B_x \in \cal B$ such that $u \in B_x \subset U$.
But then $x \leqslant u$ and in turn, $B_u \subset B$.
Now, let $v \in V$ be arbitrary. By a similar reasoning as before, we see that $v \in B_v \subset V$. Since $u \notin V$, we see that $v \not\leqslant u$. Since $X$ is totally ordered, we get $u < v$. In turn, $v \in B_u \subset U$.
Thus, $V \subset U$. $\qquad \Box$
From the claim, it follows that if $U, V \subsetneq X$ are open, then the union is equal to one of $U$ or $V$ and in turn, is not $X$.
A: I have got my own proof:
Earlier, for the $T_2$ proof, I have proved that $\mathcal B(x)=\{B_x\}$ is basis of neighbourhood of $x$ in $(X,T)$.
Now we are going to prove that there not exists two non-empty disjoint closed sets by contradiction. Suppouse that $F_1,F_2\in \mathcal C_T\setminus\{\emptyset\}$ are two non-empty closed sets with $F_1\cap F_2= \emptyset$. So, $\exists x_i\in F_i,\ i=1,2.$ As $F_1\cap F_2=\emptyset$, then $x_i\in X\setminus F_j, \ i,j=1,2, \ i\neq j$.
As $\mathcal B(x_i)=\{B_{x_i}\}$ is basis of neighbourhood of $x_i$ in $(X,T)$, then $x_i\in B_{x_i}\subset X\setminus F_j, \ i,j=1,2, \ i\neq j$.
And, as $(X,\leq)$ is totally ordered, $x_1\leq x_2$ ò $x_2\leq x_1$. Supouse that $x_1\leq x_2$ (resp. $x_2\leq x_1)$, then $x_2\in B_{x_1}\subset X\setminus F_2$!! Contradiction.
Therefore there not exists two non-empty disjoint closed sets.
