Nested
Definition. A sequence of sets $\{ I_n \}_{n=1}^\infty$ is said to be nested if $I_n \supset I_{n+1}, \forall n \in \mathbb{N}$.
Nested Interval Theorem
Theorem. If $\{ [a_n, b_n] \}_{n=1}^\infty$ is a nested sequence of intervals, then
(i) $\bigcap_{n=1}^\infty [a_n, b_n] \neq \emptyset$.
(ii) If $\lim_{n \to \infty} (b_n - a_n) = 0$, then $\bigcap_{n=1}^\infty [a_n, b_n]$ is a singleton.
Proof.
(i) Since $\{ a_n \}$ is increasing and bounded, and $\{ b_n \}$ is decreasing and bounded, $a_n \uparrow a$ and $b_n \downarrow b$ for some $a, b \in \mathbb{R}$. Since $a_n \le a \le b \le b_n, \forall n \in \mathbb{N}$, we have $$x \in [a, b] \iff x \in [a_n, b_n], \forall n \in \mathbb{N} \iff x \in \bigcap_{n=1}^\infty [a_n, b_n].$$ Thus $\bigcap_{n=1}^\infty [a_n, b_n] \neq \emptyset.$
(ii) Suppose that $\lim_{n \to \infty} (b_n - a_n) = 0$. Since $0 \le b - a \le b_n - a_n, \forall n \in \mathbb{N}$, we have $\lim_{n \to \infty} (b - a) = 0$, which implies that $a = b$, and $\bigcap_{n=1}^\infty [a_n, b_n]$ is a singleton. $\blacksquare$