Show that $\mathbb{R}$ is not a union of closed sets containing no nonempty open intervals. From Abbott's 'Understanding Analysis':

This exercise is given before Baire's theorem.
 A: I will use the Nested Interval Theorem to show it.
Because $F_n$ has no nonempty open intervals, it also has no nonempty closed intervals (except for singletons [a,a]).
Let $G_n = F_n^c$. Because $F_n$ is closed, $G_n$ is open.
Because $F_1$ has no nonempty open intervals, $\exists x_1 \in \mathbb{R}$ such that $x_1 \in G_1$. Because $G_1$ is open, $\exists \epsilon_1 > 0$ such that $V_{\epsilon_1}(x_1) \subseteq G_1$. Because $V_{\epsilon_1}(x_1)$ is an open interval, it can be expressed as $(x_1 - \epsilon_1 , x_1 + \epsilon_1)$. Then define $I_1 = [x_1 - \frac{\epsilon_1}{2}, x_1 + \frac{\epsilon_1}{2}]$. Because $F_2$ has no nonempty open intervals, $\exists x_2 \in I_1$ such that $x_2 \in G_2$. Because $G_2$ is open, $\exists \epsilon_2 > 0$ such that $V_{\epsilon_2}(x_2) \subseteq G_2$. In particular, you can take $\epsilon_2 > 0 $ such that $V_{\epsilon_2}(x_2) \subseteq I_1$.Define $I_2 = [x_2 - \frac{\epsilon_2}{2}, x_2 + \frac{\epsilon_2}{2}]$ \ In general, because $F_n$ has no nonempty open intervals, $\exists x_n \in I_{n-1}$ such that $x_n \in G_n$. Because $G_n$ is open, $\exists \epsilon_n > 0$ such that $V_{\epsilon_n}(x_n) \subseteq G_n$. In particular, you can take $\epsilon_n > 0$ such that $V_{\epsilon_n}(x_n) \subseteq I_{n-1}$. Define $ I_n = [x_n - \frac{\epsilon_n}{2}, x_n + \frac{\epsilon_n}{2}] $.
Hence , we constructed a sequence of Nested Close Intervals $ I_1 \supseteq I_2 \supseteq $... 
By the Nested Interval Theorem, $ \bigcap_n^{\infty} I_n \neq \emptyset $. Because $ I_n \subset V_{\epsilon_n}(x_n) \subseteq G_n$, we have $\bigcap_n^{\infty} G_n \neq \emptyset $.
Then, taking the complement: 
$(\bigcap_n^{\infty} G_n)^c = \bigcup_n^{\infty} F_n \neq \mathbb{R}$
