Let $U ⊂ R^n$ be an open set. Prove that there is a sequence of pairwise disjoint closed balls $\bar{B_i} ⊂ U$, $i = 1, 2, 3, . . .$ such that $|U \setminus\cup_{i=1}^∞ \bar{B_i}| = 0$. Here $U$ is the lebesgue measure(outer measure) in $\Bbb R^n$.
My first try: First I enumerated all the rationals i.e elements of $U \cap \Bbb Q^n=\{q_n\}.$ Now I consider $q_1-x_1(say)$ and chose an open ball $B(x_1,r_1) \subseteq U$, name $B(x_1,r_1)=B_1$ now $U\setminus \bar{B_1}$ is open then we will go to $x_2$ if it is in $\bar{B_1}$ we will skip it and so on chhose the next possible $q_i=x_2(say)$ such that $x_i \notin \bar{B_1}$ then $B(x_i,r_2)\subseteq U\setminus \bar{B_1}$ call $B(x_i,r_2)=B_2$ then $U\setminus \bar{B_1} \setminus \bar{B_2}$ we will move to next $q_j=x_3(say)$.
Now I was guessing that $$|U \setminus\cup_{i=1}^∞ \bar{B_i}| = 0.$$
I am unable to prove it, what I can see is if $x \in U \setminus\cup_{i=1}^∞ \bar{B_i}$ then $\not \exists$ any $\epsilon >0$ s.t $B(x, \epsilon \subseteq U \setminus\cup_{i=1}^∞ \bar{B_i}$.
My second try:
I understood that there exists and $F_{\sigma}$ set $F$ such that $|U \setminus F|=0$ so $F=\cup _{i=1}^\infty F_i$ where $F_i$'s are closed. Now how to conclude my result from here?