Let $(X,d)$ be a metric space. Suppose that exists an $r>0$ that satisfies this property:
For every $n$, there exist $x_1,\dots,x_n$ so that the balls $B_r(x_1),\dots,B_r(x_n)$ are not covering $X$ and the balls are pairwise disjoint.
Is this enough to prove that the space is not totally bounded? In a subset of $\mathbb{R}^N$ with the euclidean metric it works, but how about in a general metric space?
A metric space $(X,d)$ is totally bounded if for every $r > 0$ $X$ can be covered by finitely many balls of radius $r$. So, $X$ not totally bounded means that there exists $r>0$ so that $X$ is not covered by finitely many balls of radius $r$. Let's see what that implies. Take any point $x_1$ in $X$. The ball $B(x_1, r)$ does not cover $X$, so there exists $x_2$ so that $d(x_1, x_2)\ge r$. The balls $B(x_1, r)$, and $B(x_2, r)$ do not cover $X$, so there exists $x_3$ so that $d(x_1, x_3) \ge r$, $d(x_2, x_3)\ge r$. The construction can be continued inductively. We get a sequence $(x_n)_{n\ge 1}$ so that $d(x_i, x_j)\ge r$ for $i\ne j$ ( called an $r$-set). Conversely, if $X$ has an infinite $r$-set, then $X$ cannot be covered by finitely many balls of radius $\frac{r}{2}$, so $X$ is not totally bounded.
Notice: if $(x_n)$ is an (infinite) $r$-set then the balls $B(x_i, \frac{r}{2})$ are disjoint.
Also, if a space $X$ is totally bounded, then for every $r>0$ there exists $N_r$ so that all $r-sets$ have cardinality at most $N_r$. Indeed, $N_r$ is at most the smallest number of balls of radius $\frac{r}{2}$ that cover $r$. I guess this actually answers the question stated.
$\bf{Added:}$ A standard example of a set that is not totally bounded is the unit ball in an infinite dimensional Hilbert space. Take $(e_n)$ an infinite orthonormal basis. Then $d(e_n, e_m) = \sqrt{2}$ for all $n\ne m$. One can show that the unit ball of any infinite dimensional normed space is not totally bounded.