Cantor's Criterion: If $\left\{C_{k}\right\}$ is a nested sequence of closed bounded, nonempty subsets of $\mathbb{R}^{n}$, then $$ \bigcap_{k=1}^{\infty} C_{k} \neq \emptyset $$ Furthermore, if $\lim _{k \rightarrow \infty} d\left(C_{k}\right)=0$, where $d\left(C_{k}\right)$ is the diameter of $C_{k}$, then $\cap_{k=1}^{\infty} C_{k}=\left\{x_{0}\right\}$ for some $x_{0}$ in $\mathbb{R}^{n}$
Discussion about the Cantor's Criterion when it is not Euclidean metric: Find a counter-example and consider the statement when "closed, bounded" is replaced by "compact". Are there any examples for which Cantor's Intersection Theorem would not hold for the non-Euclidean metric? Is it related to the Heine-Borel theorem, or it should be a completely different metric?
The Cantor Intersection theorem holds in general topological spaces, so no metric notion is required. This is the topological version: Let $X$ be a topological space. A decreasing nested sequence of non-empty compact subsets of $S$ has a non-empty intersection.
What may fail, if I understand correctly, is to assume $X$ is closed and bounded and $X$ is not compact. Then you may find indeed counterexamples since the theorem stated above requires the notion of compactness. For example take, in the space of rational numbers, the sets
$$C_k = [\sqrt{2}, \sqrt{2}+1/k] = (\sqrt{2}, \sqrt{2}+1/k) $$
are closed and bounded, but their intersection is empty (as Wikipedia also states).
Note that compactness is equivalent to closed and bounded only in particular spaces, not in general, while the converse is always true (if we assume the space is not too bad, that technically means requiring it is Hausdorff, but leave this part aside).
Note: for compact I mean that for every open cover there exists a finite subcover and it satisfies the Hausdorff condition.