Gromov compactness theorem

Question

Reference: this book, page 493.

For a compact metric space $X$ define $\text{Cov}(X,\epsilon)= \min \{n \, : \, X \text{ is covered by $n$ closed } \epsilon\text{-balls} \}$ and $\text{Cap}(X,\epsilon)= \max \{n \, : \, X \text{ contains $n$ disjoint } \epsilon/2\text{-balls}\} $. Then we have the following

Gromov compactness theorem. Let $\mathcal{C} \subset \mathcal{M}$ (the class of all compact metric spaces) be a class of compact metric spaces. The following are equivalent:

1. $\mathcal{C}$ is precompact, i.e. every sequence in $\mathcal{C}$ contains a subsequence which converges in $\mathcal{M}$ (in the Gromov-Hausdorff metric).
2. There exists a function $N(\epsilon) : (0, \beta) \to (0, \infty)$ such that $\text{Cap}(X,\epsilon) \le N(\epsilon)$ for all $\epsilon \in (0, \beta)$ and $X \in \mathcal{C}$.
3. There exists a function $N(\epsilon) : (0, \beta) \to (0, \infty)$ such that $\text{Cov}(X,\epsilon) \le N(\epsilon)$ for all $\epsilon \in (0, \beta)$ and $X \in \mathcal{C}$.

Now, let's put $X_n = \{1,n\}$. For every $\epsilon > 0$ we have that $B_{\epsilon}(1) = \{x \in X_n \, : \, |x-1| < \epsilon \} = \{1\}$ and that $B_{\epsilon}(n) = \{x \in X_n \, : \, |x-n| < \epsilon \} = \{n\}$; therefore $N(\epsilon) = 2$ for every $X_n$, but clearly $\{X_n\}_{n \in \mathbb{N}}$ doesn't admit a convergent subsequence.

It seems to me that a hypothesis is missing - i.e. a "uniform control" on the diameters of every set in $\mathcal{C}$. This is the second textbook in which I've found this (wrong?) version of the theorem (in the Gromov's original paper, for example, Gromov uses the uniform compactness).

Am I wrong?

Thank you in advance.