Discrete family of compact subsets in a metric space A family $\{A_i: i \in A_i\}$  of subsets of a topological space $X$ is said to be discrete if each $x \in X$ has a neighborhood meeting at the most one $A_i$.
Suppose $(X,d)$ is a metric space and $\{A_n: n \in \mathbb{N}\}$ a discrete family of compact subsets of $X$. How can we show that there exists a sequence $\varepsilon_{n}$ of positive number which converges to $0$ and the family $\{B(A_n,\varepsilon_{n}): n \in \mathbb{N}\}$ is a discrete family. Where $B(A_n,\varepsilon_{n})$ denotes the union of all balls centered at the points of $A_n$ and radius $\varepsilon_n$. 
 A: This seems false:
In $\mathbb R$, choose the family $A_1=[-1,0]$ and $A_n=[2^{-2n-2},2^{-2n-1}]$ for $n\geq2$.  No matter what $\varepsilon_1>0$ we choose, there will be some $x\in B(A_1,\varepsilon_1)$ which is also in $A_n$ for some $n$.
A: For every $n\in\mathbb N$, there exists a $\delta_n$ such that $B(A_n,\delta_n)$ is disjoint from every $A_m$, with $m\neq n$.  To get $\delta_n$, cover $A_n$ with neighborhoods around each point which intersect only $A_n$, using the discreteness of that family.  The union of these neighborhoods is an open set containing $A_n$, so the complement is a closed set, call $C_n$.  The distance from a compact set ($A_n$) to a disjoint closed set ($C_n$) is positive, so call this distance $2\delta_n$.  Now to ensure the disjointness of the neighborhoods, we need to replace $A_n$ with the compact set $\operatorname{Cl}\left(B(A_n,\delta_n)\right)$ before we choose $\delta_{n+1}$, and so on...  Finally, to get the limit to go to zero, we choose $\varepsilon_n=\min\{\delta_n,1/n\}$.
