From Zorich, Mathematical Analysis II, pag. 16:
I believe that lemma 4 is wrongly stated, since if we were actually thinking of a metric space $(K, d)$ with $K$ compact (i.e. such that from any cover of open sets according to the topology induced in $K$ by the metric $d$, it is always possible to extract a finished subcover), I could very well take the following counterexample that violates the proposition:
a set $ K $ consisting of only three real numbers (using the standard metric in $ \mathbb {R} $) $ \Rightarrow K $ is obviously compact but it is equally obvious that the space $ (K, d) $ does not respect the sentence above.
Am I wrong?
Yes, you are wrong. If $(K,d)$ is a finite metric space, then, for every $\varepsilon>0$ small enough, $K$ itself is a finite $\varepsilon$-grid of $K$.