Let $(X,d)$ be a metric space and define for $B\subset X$ bounded, i.e.
$\operatorname{diam}(B)= \sup \{ d(x,y) \colon x,y\in B \} < \infty,$
the measure
$\beta(B) = \inf\{r > 0\colon\text{there exist finitely many balls of radius r which cover } B\},$
or equivalently,
$\beta(B)=\inf\big\lbrace r > 0|\exists N=N(r)\in{\bf N} \text{ and } x_1,\ldots x_N\in X\colon B\subset\bigcup_{k=1}^N B(x_k,r)\big\rbrace,$
where $B(x,r)=\{y\in X\colon d(x,y)\}$ denotes the open ball of radius $r$ centered at $x\in X$. Let ${\bf K}(X)$ denote the collection of (non-empty) compact subsets in $X$.
I would like to prove
$\beta(B)=d_H\big(B,{\bf K}(X)\big),$ where $d_H$ is the Hausdorff distance.
I proved $d_H\big(B,{\bf K}(X)\big)\le\beta(B)$. Is there someone that knows how to prove the other inequality?