Intersection of open ball under compact set Let $(E,d)$ a metric space, and a compact set $K$ and $r>0$
I want to prove that  $U=\cap_{x\in K} B(x,r)$ is open in $(E,d)$
Using sequences i proved that $E\setminus U$ is closed then $U$ is open.
Now i want to prove this by this way: let $y\in U$ then $\forall x\in K, d(x,y)<r$ 
and i must find $\rho>0$ such that $B(y,\rho)\subset \cap_{x\in K}B(x,r)$
i take $\rho=r-d(x,y)$ but i don't see the fact that $K$ is compact with this methods .
Can someone help me.
Thank you 

Suppose $u \in U$ and let $\rho = \max_{k \in K} d(u,k)$. This is well
defined since $K$ is compact and $k \mapsto d(u,k)$ is continuous. We must have 
$\rho < r$
and
so
if $y \in B(u,r-\rho)$ and $k \in K$, we have 
$d(y,k)\le d(y,u)+d(u,k) < r -\rho+ \rho = r$, and so $B(u,r-\rho) \subset U$.
