This can be generalized:

Let $(C_n)$ be a sequence of sets of $\mathbb{R}^m$ such that $\sum_n \text{diam}(C_n)^m=S<\infty$. Then we can find disjoint sets $(D_n)_n$ such that $D_n$ is a translate of $C_n$ and $\bigcup_{n=1}^\infty D_n$ is bounded.

To prove that, reorder the sets $C_n$ so that the sequence $d_n:=\text{diam}(C_n)$ is decreasing. Now consider a ball $B\subseteq\mathbb{R}^m$ of volume $4^m S$. Then we can find by recursion a sequence of points $p_n\in B$ such that for any $i<j$ we have $d(p_i,p_j)>2d_i$. To do this we just need to choose $p_n\in B$ outside the closed balls $B(p_1,2d_1),B(p_2,2d_2),\dots,B(p_n,2d_n)$: this is possible because, if $m$ is Lebesgue measure, $m(B)=4^mS=\sum_{i=1}^\infty(2\text{diam}(C_i))^m\geq\sum_{i=1}^{n-1}(4\text{diam}(C_i)^m)>\sum_{i=1}^{n-1}m(B(p_i,2d_i))$.

Then choose $D_n$ to be some translate of $C_n$ containing $p_n$. Clearly $\bigcup_n D_n$ is bounded, and the $D_i$ are disjoint because for any $i<j$, $d(p_i,p_j)>2\text{diam}(C_i)\geq \text{diam}(C_i)+\text{diam}(C_j)$.