It is a Hausdorff, therefore, compact implies closed. If it were not bounded, then for any ball $B(x,n)$ there would be a point outside it, and while the union of all these balls does cover the whole space, there is no finite subcovering for the subspace. See also the proof of Theorem 27.3. Now we must find an example of a space where there is a closed bounded set such that it is not compact. Boundness is not a topological property. Therefore, we can take any non-compact metric space and make it bounded by taking the standard bounded metric. Or, for example, take an infinite space in the discrete topology (with the discrete metric: $d(x,y)=1$ for $x\neq y$ ).