Let $\{ B_1, ..., B_k \}$ be a collection of balls in $\mathbb{R}^n$. Then there exists a finite sub collection $\{ B_{j_1}, ..., B_{j_{\ell}} \}$ of pairwise disjoint balls such that $$\sum_{r=1}^{\ell} \left| B_{j_r} \right| \geq 3^{-n} \left| \bigcup_{i=1}^k B_i \right|.$$
To prove this we order the $B_i$ by its Lebesgue measure. That is $\left| B_1 \right| \geq \left| B_2 \right| \geq \cdots \geq \left| B_k \right|$ and choose $j_1 =1$. We then form the pairwise disjoint collection iteratively by choosing the smallest index $s$ such that $B_s$ is disjoint from $\bigcup_{m=1}^i B_{j_m}$. This gives us our collection $B_{j_1}, ..., B_{j_{\ell}}$. Now suppose $B_m$ is not in this collection. Then $m$ must intersect some $B_{j_r}$ for some $j_r <m$. By considering the ordering that we defined earlier it follows that $\left| B_m \right| \leq \left| B_{j_r} \right|$.
The claim that follows is that $$B_m \subseteq 3B_{j_r},$$ which is the crux of the proof. I don't see how we can ensure this.
This is geometry. The facts that $B_m$ intersects $B_{j_r}$ and $|B_m| \le |B_{j_r}|$ imply $B_m \subseteq 3B_{j_r}$. Draw a picture for $n=2$ maybe.