Bounded Metric Space is not necessarily Totally Bounded/Proof 2

Proof
Let $d$ be a discrete metric on the open unit interval $\Bbb I := \openint 0 1 \subseteq \R$.

We have that for all $x \in \openint 0 1$ and for all $r \in \R_{> 1}$:
 * $\map {B_r} x = \openint 0 1$

where $\map {B_r} x$ denotes the open $r$-ball of $x$.

Thus $\struct {\Bbb I, d}$ is bounded.

Let $\epsilon \in \R_{>0}$ be a strictly positive real number such that $\epsilon < 1$.

Let $x \in \openint 0 1$ be arbitrary.

Then $\map {B_\epsilon} x$ contains $x$ alone.

So, there is no finite $\epsilon$-net for $\openint 0 1$.

By definition of totally bounded metric space, $\struct {\Bbb I, d}$ is not totally bounded.