Does not exist cover of $\mathbb{R}^n$ by disjoint closed balls 
Does not exist cover of $\mathbb{R}^n$ by disjoint closed balls with positive radius.

My attempt: Suppose that exists, we can write:
$\mathbb{R}^n=\displaystyle\bigcup_{i=1}^{\infty} B_{i}$. Let $C$ denote the set of center points of these balls. All points of $C$ are isolated so $C$ is countable. Associating each point of $C$ to the ball of wich this point is the center,it is easy see that this association is a bijection, concluding that set of balls is countable.
The set of limit points of $C$, $C^{'}=\displaystyle\bigcup_{i=1}^{\infty} \partial B_{i}$. I thought it would be easy to find a contradiction there, I was wrong. Until now, I don't use hypothesis that $B_{i}$ is closed, and I think that is it that lack in my demostration.
Is it true if the balls are open?
Thank you for any help.
 A: For the Baire category proof.
Use the OP proof up to
$$
C^{'}=\displaystyle\bigcup_{i=1}^{\infty} \partial B_{i} .
$$
Then note $C^{'}$ is a complete metric space, written as a countable union of sets $\partial B_i$, which are closed (in $C'$) sets with empty interior (in $C'$).
A: To show no such cover via open balls exists is easy: any such cover $\mathcal{O}$ must involve at least 2 open balls (since the radii are finite). So pick $O\in\mathcal{O}$, and let


*

*$A=O$,

*$B=\bigcup (\mathcal{O}\setminus\{O\})$.
Then $A$ is open by assumption, while $B$ is open as a union of open sets; and $A\sqcup B=\mathbb{R}^n$. But this contradicts the connectivity of $\mathbb{R}^n$.

Here's one way to do it. Suppose towards contradiction that $\mathcal{U}$ is a set of countably many disjoint closed balls with finite positive radius, which covers $\mathbb{R}^n$  (you've justified in your question why we may assume the cover is countable). Say that a closed ball $B$ is shattered if it is not contained in a single element of $\mathcal{U}$.


*

*Exercise: there is a shattered closed ball.


Now the crucial point is:

If $B$ is a shattered closed ball, and $U\in\mathcal{U}$, then there is a closed ball $C$ of aribtrarily small positive radius contained in $B\setminus U$ which is also shattered. 

Why? Well, clearly there are $x, y\in B\setminus U$ which lie in different elements $V_x, V_y$ of $\mathcal{U}$ (why?). This means that $\partial V_x\cap (B\setminus U)$ is nonempty (why?); let $z$ be an element of this boundary. Then an appropriately small-radius closed ball around $z$ is shattered.
OK, so what? Well, now we can diagonalize against $\mathcal{U}$! Let $\mathcal{U}=\{U_i: i\in\mathbb{N}\}$, and define a sequence of shattered closed balls of finite positive radius $C_i$ such that


*

*$C_{i+1}\subseteq C_{i}\setminus U_i$, and

*The diameter of $C_i$ is at most $2^{-i}$.
But then $\bigcap C_i=\{\alpha\}$ for some $\alpha\in\mathbb{R}^n$, which can't be an element of any element of $\mathcal{U}$ (why?); contradiction.
