Proving non compactness of a space

Question

I'm trying to show that the space $\mathbb R^p$, endowed with a metric $d'(x,y) = \frac{d_2(x,y)}{1 + d_2(x,y)}$, where $d_2(x,y)$ is the Euclidean distance, is closed and bounded but not compact.

I've had no problem with the first two proof, but I cannot go ahead with the proof of non compactness. I only know that I have to use the Bolzano-Weierstrass property about subsequences and to proceed by contradiction, assuming that the space is compact.

Answer 1

Other answer using sequences.

The sequence $(a_n)$ with $a_n=(n,0, \dots, 0)$ is bounded (by $1$) but has no converging subsequence. Hence applying Bolzano–Weierstrass theorem, our metric space is not compact.

Answer 2

The topology induced by $d'$ is the same as the standard topology and you probably know $\Bbb R^p$ is not compact. Or directly: As $d(0,x)$ can get arbitrarily close to $1$ but never euqal, we have $\Bbb R^p=\bigcup_{0<r<1}B_{d'}(0;r)$ but no finite subcover suffices.