Exercise 13.7

The problem is very abstract so it may be a bit confusing when one starts trying to solve it. Here is my solution.

Proof: Let consider an open set $U \subset \mathbb{R}$. Notice that, $U$ may be a union of infinite open sets, which is also an open set. Now, let consider $A = (U \cap \mathbb{Q})$, i.e. $A$ is the set of all rational numbers in $U$. Let $a_n = \inf \{ a \in \mathbb{R} : (a, q_n] \subset U \}$ and $b_n = \sup \{ b \in \mathbb{R} : [q_n, b) \subset U   \}$ where $q_n \in A$. Then $\cup_{i=1}^{\infty} (a_n, b_n) = U$.
We will prove $(a_n, b_n) \cap (a_m, b_m) \neq \emptyset \implies (a_n, b_n) = (a_m, b_m)$ as follows.

Suppose $(a_n, b_n) \cap (a_m, b_m) \neq \emptyset$, Let $k \in (a_n, b_n) \cap (a_m, b_m)$. Because $k \in (a_n, b_n) \cap (a_m, b_m)$ and $ (a_n, b_n) \subset U$, $(a_m, b_m) \subset U$. By completeness axiom, $\inf \{ a \in \mathbb{R} : (a, k] \subset U \}$ is unique, therefore $a_n = a_m$. Analogously, $\sup \{ b \in \mathbb{R}: [k,b) \subset U \}$ is unique, therefore $b_n = b_m$ and we are done.

A bit more about the last sentence: $\{ a \in \mathbb{R} : (a, k] \subset U \} \subset \mathbb{R}$ is either bounded below or unbounded below. If it is bounded below, then by the completeness axiom, it has a unique infimum. Otherwise, its "infimum" is $-\infty$. Similarly, $\{ b \in \mathbb{R} : [k,b) \subset U \} \subset \mathbb{R}$ and it is either bounded above or unbounded above. If it's bounded above, then it has a unique supremum, otherwise, its "supremum" would be $+\infty$. No matter which condition we are on, the "supremum" and "infimum" are two unique numbers or $+\infty$ and $-\infty$, respectively. Therefore, $\sup \{ b \in \mathbb{R} : [k,b) \subset U \} $ is unique and so is the $\inf \{ a \in \mathbb{R} : (a, k] \subset U \}$.