The version for metric spaces of Cantor's intersection theorem says that if $(C_n)_{n \ge 0}$ are nonempty nested closed sets of a complete metric space $(X,d)$, with $\operatorname{diam } C_n \to 0$, then $\bigcap_{n \ge 0} C_n$ consists of a single point. The proof is short and elementary.
What happens if we relax the diameter condition to say that all the $C_n$ are bounded and that $\operatorname{diam } C_{n+1} \le \operatorname{diam } C_n$? More clearly, is it possible to have $\bigcap_{n \ge 0} C_n = \emptyset$? Intuition says no, but I cannot find a proof.
Example or Counterexample: $(X, d) = (\{e_1, e_2, \cdots\}, d_{dis})$ with the discrete metric $d_{dis}$. Then all subsets of $X$ are closed and bounded. Let
$$C_n = \{ e_n, e_{n+1}, \cdots\},$$
Indeed, the condition on the diameter is trivial since $C_n$ is a nested sequence.