Let $X$ denote a complete metric space. Suppose $\mathcal{C}$ is a collection of closed and bounded subsets of $X$ with the finite intersection property.
Question. Is $\bigcap\mathcal{C}$ necessarily non-empty?
A couple of remarks.
Firstly:
To see that closedness is necessary, consider the sequence of open intervals $(0,1/n)$ as subsets of the real line.
To see that boundedness is necessary, consider the sequence of closed intervals $[n,\infty)$ as subsets of the real line.
Secondly: I'm also interested in the converse; that is, whether the above condition implies completeness. If so, this gives us a notion of completeness for any set $X$ equipped with both a topology and a bornology.
So, it turns out there's a really easy counter-example. Put a metric space structure on $\mathbb{N}$ by declaring that the distance between any two distinct points is exactly $1$. This is complete, because the only Cauchy-sequences are the constant sequences, which are obviously convergent. Also, every subset of $\mathbb{N}$ has diameter at most $1$, and is therefore bounded; further, the induced topology is the discrete topology, so every subset is closed. So in this case, the question reads:
Of course, the answer is "no": consider, for example, $$\mathcal{C} = \{[n,\infty)_{\mathbb{N}} : n \in \mathbb{N}\}.$$