TITLE: Filling a unit cube with countable balls.
QUESTION [9 upvotes]: Is it possible to fill an $n-$dimensional unit cube with countable number of non-overlapping $n-$dimensional balls? 

By $n-$dimensional unit cube, I am thinking of the set
$$
C:=\{\ x\in\Bbb R^n\ |\ 0\le x_i\le 1\ \text{for all}\ i=1,2,\dots,n\ \}
$$
where $x=(x_1,\dots,x_n)$. An $n-$dimensional ball centered at $x_0$ is defined to be
$$
B(x_0;r):=\{\ x\in\Bbb R^n\ |\ d(x_0,x)<r\ \}
$$ 
where $d$ is the Euclidean distance function.

A sequence of balls $(B_1,B_2,\dots)$ is said to fill the unit cube if they are pairwise-disjoint, each $B_i\subset C$ and
  $$
\sum_{i=1}^{\infty}\mu(B_i)=1
$$ 
  where $\mu$ is the Lebesgue measure on $\Bbb R^n$.

Intuitively, I feel that it should be possible to fill the unit cube with many balls. However, I can't think of a way to prove it. 
Since $C$ is separable, I thought about enumerating its rational points into a sequence $\{c_1,c_2,c_3,\dots \}$ and form balls around them such that $B_i:=B(c_i,r_i)$ where
$$
r_i:=\sup\{\ r\in\Bbb R^+\ |\ B(c_i;r)\ \text{is contained in $C$ and doesn't intersect $B_1,B_2,\dots,B_{i-1}$} \}.
$$ 
I don't know if this method works or not. I would really appreciate if anyone can suggest a way to finish the prove, suggest a new method, or give a negative answer to my question. An answer that only works in dimension $2$ or $3$ is also very welcomed.

REPLY [3 votes]: I'd like to add some remarks besides the elegant proof by Eric Wofsey.
Vitali's covering theorem implies that not only you can do it, but also that you can use balls from any given family $\mathcal{V}=\{B_j\}_{j\in J}$, provided it is a Vitali covering of your set (this means that any point $x\in C$ is contained in balls with arbitrarily small radius).
Moreover the same is true with minor modifications for any Radon measure, if you consider a family of closed balls (obviously this is irrelevant for the Lebesgue measure). To prove this result however you need a different covering theorem, namely Besicovitch's. You can see for instance the first chapter of the book by Evans and Gariepy, Measure Theory and Fine Properties of Functions.
The first linked Wikipedia article also states a version with balls replaced by any family of sets which are not "too thin", meaning that
$$\mathrm{diam}(B)^n\leq C \mathcal{L}^n(B)$$ 
for any set $B$ in the family (this is closely related to the last paragraph in the answer by Eric).
It would be really interesting to know if your proposed algorithm works. I've thought about it some time but I didn't go anywhere for now.