I know it is true that on the real line that $\forall B$, Borel measurable set with finite measure, $\forall \varepsilon>0$ there exists disjoint countable open balls $B_i$ cover $B$ with $$ \mu (\cup B_i)\leq \mu(B)+\varepsilon ,$$ where $\mu$ is the Lebesgue measure.
This is proven by first noting that we have $U$ open, $B\subset U$ and $$\mu(U)\leq \mu(B)+\varepsilon.$$
Then using the fact that on the real line, we could decompose every open set to countable disjoint open intervals(balls).
Would the same hold for $\mathbb{R}^n,n\geq 2$?
No, this is not true in higher dimension. This is because only in one-dimensional case it holds that every open set is an union of disjoint balls.
For a counterexample, consider the interval $[0,1]\times \{0\}$. If it's covered by a family of disjoint balls, it is covered by some finitely many of them. But then it's easy to see that it must be covered by just one of them, and its measure is at least $\pi/4$ (because the interval can't be smaller than the diameter).