Lemma. Let $C$ be a collection of open balls in $\mathbb{R}^n$, and let $U =\cup_{B\in C} B$. If $c<m(U)$, there exist disjoint $B_1,...,B_k\in C$ such that $\sum_{k=1}^{k}m(B_j)> 3^{-n}c$.
Here $m$ is the lebesgue measure.
In the proof, it first picks up one compact $K\subset U$ with $m(K)>c$ and finitely many of the balls $A_1,...,A_p$ in $C$ to cover $K$. Let $B_1$ be the largest of the $A_j$'s (which have the maximal radius), let $B_2$ be the largest of the $A_j$'s that are disjoint from $B_1$, and so on until the list of $A_j$'s is exhausted after $k$ steps. If $B_i = B(x_i,r_i)$, put $B_i^*=B(x_i,3r_i)$. Then $K\subset \cup_{i=1}^{k}B_i^*$ and $c<\sum_{i=1}^k m(B_i^*) = 3^n\sum_{i=1}^k m(B_i)$.
Question: Suppose $C$ contains finitely many balls which are intersected with each other. Then how to do this construction.