Proving that order topology on a well-order is a normal space I don't understand something about the following proof:
"Let $(X,\le)$ a well-ordered set. Then $X$ is a normal space with the order topology.
Let $F_1,F_2\subseteq X$ closed sets such that $F_1\cap F_2=\emptyset$. First, suposse $\min X\notin F_1\cup F_2$. Thus, for each $y\in F_1\subseteq X\setminus F_2$ we can find $x_y\in X$ such that $(x_y,y]\subseteq X\setminus F_2$.
Similarly, for each $b\in F_2$, we can find $a_b\in X$ such that $(a_b,b]\subseteq X\setminus F_1$.
Consider $U=\displaystyle\bigcup_{y\in F_1}(x_y,y]$ and $V=\displaystyle\bigcup_{b\in F_2}(a_b,b]$, then $U\cap V=\emptyset $..."
I don't see why $U\cap V=\emptyset$... Can anyone explain this to me?
Thanks.

Suppose that $z\in U\cap V$. Then there must be a $y\in F_1$ and a $b\in F_2$ such that $z\in(x_y,y]$ and $z\in(a_b,b]$. This means that $x_y<z\le y$ and $a_b<z\le b$. Since $y\in F_1$ and $b\in F_2$, and $F_1\cap F_2=\varnothing$, it’s clear that $y\ne b$, so either $y<b$, or $b<y$. Suppose that $y<b$; the other case is completely similar. Then $a_b<z\le y<b$, so $y\in(a_b,b]$. 
OOPS: we chose $a_b$ so that $(a_b,b]\cap F_1=\varnothing$, and now we have $y\in(a_b,b]\cap F_1$. That’s impossible so no such $z$ can exist, and $U\cap V$ must be empty.
