A example of closed and bounded does not imply compactnesss in metric Space

Let $X$ be the integers with metric $ρ(m,n)=1$, except that $ρ(n,n)=0$. Check that $ρ$ is a metric. Show that $X$ is closed and bounded, but not compact.

This is a "made-up" example demonstrating closed and bounded doesn't imply compactess in more general metric space. I checked that $\rho$ is a metric already. Yet I have no idea how to approach "showing $X$ is closed and bounded." I visualized this metric to be a set of number with just $0$ and $1$ (or maybe this is not correct?). Also, I doubt that this metric is NOT compact. Anyway, I'd appreciate if you can help! Meanwhile, do we use a ball $B(0,1)$ in general to show that a metric is closed and bounded? If so, why?

Your metric generates the discrete space where every subset of $X$ is open (and thus also closed). It is bounded, because each point lies within a distance $1$ of some point $x_0$ (any will do). It is not compact, because $\{\{x\} \mid x \in X\}$ is an open cover of $X$, but you won't be able to pick finite subcover, because $X$ is infinite.

I hope this helps ;-)

Here is a simple example. Denote by $\ell^\infty$ the set of all bounded sequences of real numbers; put $$d(x,y) = \sup_n |x_n - y_n|.$$ Then all sequences of distance $\le 1$ from the zero sequence is closed, bounded but it is not compact.