why compactness of metric space K means it can be covered by finite small open ball $d(x,p)<\delta$ In the process to prove theorem 7.25 of the Principles of Mathematical Analysis, I can't exactly understand why 
"Since K is compact, there are finitely many points $p_1,...,p_r$ in K such that to every $x \in K$ corresponds at least one $p_i$ with $d(x,p_i)<\delta$"
is valid,
I understand the definition of compactness in chapther 2 of Rudin, and I can "heuristically" understand above statement. But I want to know the exact proof of above statement. Thank you!

If $K$ is a subset in a metric space and $\delta > 0$, then
$$
K \subset \bigcup_{x \in K} B(x, \delta)
$$
where $B(x,\delta)$ is the open ball centered at $x$ with radius $\delta$. This union is an open cover of $K$. If $K$ is compact, there exists a finite subcover
$$
K \subset \bigcup_{i = 1,\cdots,n} B(p_i, \delta)
$$
