The set $\mathbb{R}$ of real numbers with the lower limit topology having as a basis the collection of all closed-open intervals $[a, b)$, where $a, b \in \mathbb{R}$ with $a < b$, is denoted by $\mathbb{R}_l$.
Here is Example 2, Sec. 31, in the book Topology by James R. Munrkres, 2nd edition:
The space $\mathbb{R}_l$ is normal. It is immediate that one-point sets are closed in $\mathbb{R}_l$, since the topology of $\mathbb{R}_l$ is finer than that of $\mathbb{R}$. To check normality, suppose that $A$ and $B$ are disjoint closed sets in $\mathbb{R}_l$. For each point $a$ of $A$ choose a basis element $\left[ a, x_a \right)$ not intersecting $B$; and for each point $b$ of $B$ choose a basis element $\left[ b, x_b \right)$ not intersecting $A$. The open sets $$ U = \bigcup_{a \in A} \left[ a, x_a \right) \qquad \mbox{ and } \qquad V = \bigcup_{b \in B} \left[ b, x_b \right) $$ are disjoint open sets about $A$ and $B$, respectively.
In the above proof, how do we know that the sets $U$ and $V$ are indeed disjoint?
My Attempt:
Suppose that $U$ and $V$ are not disjoint. Let $p$ be a point of $U \cap V$. Then there are some points $a \in A$ and $b \in B$ such that $$ p \in \left[a , x_a \right) \qquad \mbox{ and } \qquad p \in \left[ b, x_b \right). $$ But as the interval $\left[a , x_a \right)$ does not intersect $B$ and as the interval $\left[ b, x_b \right)$ does not intersect $A$, so we must also have $$ p \in \left(a , x_a \right) \qquad \mbox{ and } \qquad p \in \left( b, x_b \right). \tag{1} $$
What next? How to proceed from here to arrive at our desired contradiction? Or, is there some alternative way of showing these sets $U$ and $V$ to be disjoint?
PS:
I think I've now managed to figure this out.
Since the interval $\left[a , x_a \right)$ does not intersect $B$, the interval $\left[ b, x_b \right)$ does not intersect $A$, $a \in A$, and $b \in B$, therefore we must have the following: Either $x_a \leq b$ or $x_b \leq a$.
If $x_a \leq b$, then we have $$ a < x_a \leq b < x_b. $$
On the other hand, if $x_b \leq a$, then we have $$ b < x_b \leq a < x_a. $$
In either case, the intervals $\left[a , x_a \right)$ and $\left[ b, x_b \right)$ will turn out to be disjoint, contrary to (1) above.
Is this reasoning correct? If so, then is it also free-of-gaps? Or, are there any other ways of correcting / improving it?
If for all x > a, [a,x) intersectes B,
then since B is closed, a is in B.
Thus an x can be chosen so [a,x) misses B.