The proof of the Baire category theorem for complete metric spaces involves, among other things, constructing a sequence $(F_n)_n$ of closed sets such that $\operatorname{diam} F_n \to 0$. This gives you a point in $\bigcap_n F_n$, concluding the proof.
Why do I have to have closed sets of shrinking diameter? Making the size of the sets small enough to obtain a point in all of them seems a bit counterintuitive. It is easy to see that I want the closed sets to be bounded, since if I take, for instance, $F_n := [n, \infty)$ in the complete metric space $\mathbb R$, then the intersection is empty. However, I cannot see why I need to have the diameters of the closed sets to be really small in the limit and tend to zero.
Question: what is an example of a sequence of closed sets in a completely metric space, every set of which is bounded, such that the intersection of all the closed sets is empty?
Let $X=\Bbb R\setminus\Bbb Q$. $X$ is not complete in the usual metric, but it’s a $G_\delta$ in the complete metric space $\Bbb R$, so it’s completely metrizable: there is a metric $d$ on $X$ such that $\langle X,d\rangle$ is complete, and $d$ generates the usual topology. Let $q\in\Bbb Q$, and let $\langle q_n:n\in\Bbb N\rangle$ be a strictly increasing sequence of rational numbers converging to $q$ in $\Bbb R$. For $n\in\Bbb N$ let $I_n=(q_n,q)\cap X$; each $I_n$ is clopen in $X$, and $I_{n+1}\subsetneqq I_n$ for each $n\in\Bbb N$, but $\bigcap_{n\in\Bbb N}I_n=\varnothing$.
In fact $X$ is well-known to be homeomorphic to $\Bbb N^{\Bbb N}$ with the product topology, where $\Bbb N$ is given the discrete topology. If $d$ is the discrete metric on $\Bbb N$, a complete metric $\rho$ on $\Bbb N^{\Bbb N}$ is given by
$$\rho(x,y)=\sum_{n\in\Bbb N}\frac{d(x_n,y_n)}{2^n}\;,$$
where $x=\langle x_n:n\in\Bbb N\rangle,y=\langle y_n:n\in\Bbb N\rangle\in\Bbb N^{\Bbb N}$.