I am trying to prove the following: If $\{A_n\}_{n=1}^{\infty}$ is a nested sequence of sets in $\mathbb{R}$ each of which is closed, bounded, and non-empty, then $\cap_{n=1}^{\infty} A_n$ is non-empty.
I know the nested interval theorem, so I tried to make progress as follows. Let $a_n = \inf A_n$ and $b_n = \sup A_n$, which are well-defined since each $A_n$ is non-empty and bounded. The intervals $[a_n, b_n]$ are a nested sequence of non-empty, closed intervals and as such $\cap_{n=1} [a_n, b_n]$ is not empty. Also, $\cap_{n=1}^{\infty} A_n \subset \cap_{n=1} [a_n, b_n]$.
However, it is unclear if I can make further progress in this direction. Simple counterexamples show that $\cap_{n=1}^{\infty} A_n \neq \cap_{n=1} [a_n, b_n]$, e.g., $A_n = [0, 1/n] \cup [1, (n+1)/n]$ whence $\cap_{n=1}^{\infty} A_n = \{0, 1\}$, while $\cap_{n=1} [a_n, b_n] = [0, 1]$.
I have noticed some other things while trying to prove this statement. For example, if $A_n$ is a finite set for any $n$, say $A_n = \{a_1, …, a_r\}$, then there must be some $i = 1,…,r$ such that $a_i \in A_m$ for all $m \geq n$ (because the sequence is nested and each set in the sequence is non-empty). Thus $\cap_{n=1}^{\infty} A_n$ contains $a_i$ and is not empty as desired.
Also, if $a$ is a limit point of $A_n$, then for $m \geq n$ either $a$ is in $A_m$ or $a$ is not a limit point of $A_m$ (since $A_m$ is closed).
Best Answer
You can make progress. You don't need to consider endpoints. What I would do is choose $x_n\in A_n$ for all $n$. Since the $A_1$ is bounded and $x_n\in A_n\subseteq A_1$ for all $n$ it follows that $\{x_n\}$ is bounded. It now follows from the Bolzano-Weierstrass property that $\{x_n\}$ has at least one convergent subsequence, let $x$ be a limit of one such subsequence. Since $x$ is a limit of a subsequence of$\{x_n:n\in\mathbb{N}\}$ it follows that it is a limit of a subsequence of $$\{x_m:m\in\mathbb{N} \ \text{and } m\geq n\}\subseteq A_n$$ for all $n$. Well then as $A_n$ is closed and $x$ is a limit of a sequence in $A_n$ we have that $x\in A_n$. Since $n$ was arbitrary $x\in \bigcap_{n\in\mathbb{N}} A_n$. Thus the set is not-empty.