I am currently working through Measure, Integration and Real Analysis by Sheldon Axler (Open Access Link). On page 24, ex. 12 states:
Suppose $r_1,r_2,\dots$ is a sequence that contains every rational number. Let
$$ F= \mathbb R \setminus \bigcup_{k=1}^\infty \left(r_k-\frac {1}{2^k},r_k+\frac {1}{2^k} \right). $$
(a) Show that $F$ is a closed subset of $\mathbb R$.
(b) Prove that if $I$ is an interval contained in $F$, then $I$ contains at most one element.
(c) Prove that $|F| = \infty$.
My solutions to (a) and (c) are as follows:
(a) Let $G=\bigcup_{k=1}^\infty \left(r_k-\frac {1}{2^k},r_k+\frac {1}{2^k} \right)$. Then $G$ is a union of open sets and thus open. $F$ is the complement of an open set and thus $F$ is closed.
(c) From ex. 3 we know that with $A,B \subset \mathbb R$ and $|A| < \infty$, then $|B\setminus A |\geq |B|-|A|$. From the countable subadditivity of outer measure it follows that
$$|G| \leq \sum_{k=1}^\infty \left|\left(r_k-\frac {1}{2^k},r_k+\frac {1}{2^k} \right)\right| = 2 < \infty.$$
So
$$|F| = |\mathbb R \setminus G| \geq \infty - 2 = \infty.$$
And we conclude that $|F| = \infty$.
Question
My question is how to go about question (b): It is known that there is a rational between any two irrational numbers. Can I argue that $F$ contains only irrationals (because we removed the rationals) and so every interval can contain at most one element?
So, to try: Suppose there is an interval $I$ containing more than one element in $F$. Clearly the interval contains only isolated irrational numbers. Take any two elements $y_1,y_2$ with $y_1 < y_2$ from $F$. Then the interval $(y_1,y_2)$ does not contain all elements $y_1 < x < y_2$ with $x \in \mathbb R$ and so is not an interval. Because the choice of $y_1,y_2$ was arbritrary we conclude that there is no interval with more than one element in $F$.
Thank you so much in advance!