Bartle, Sherbert - Introduction to Real Analysis, page 64

If $I_n=[a_n,b_n]$ is a nested sequence of closed bounded intervals, then there exists a real number $\xi$ such that $\xi \in I_n$ for every $n\in \mathbb{N}$.

Proof

Since the intervals are nested, $I_n\subseteq I_1$ for all $n\in \mathbb{N}$.
Further, $a_n\le b_1$ for all $n\in \mathbb{N}$.
Then $\{a_n:n\in \mathbb{N}\}$ is bounded above, let $\xi$ be its supremum.
Therefore, $a_n\le \xi$ for all $n\in \mathbb{N}$.

For any $n\in \mathbb{N}$, $b_n$ is an upper bound for $\{a_k:k\in \mathbb{N}\}$:

1. if $n\le k$, since $I_n\supseteq I_k$, then $a_k\le b_k\le b_n$
2. if $n>k$, since $I_n\subseteq I_k$, then $a_k\le a_n\le b_n$.

and since $\xi$ the least upper bound, $\xi \le b_n$.

Therefore, $a_n\le \xi \le b_n$ for all $n\in \mathbb{N}$, so $\xi \in I_n$ for all $n\in \mathbb{N}$.

QED