Ok, let's use the following definitions:
Definition: If $B \subset \mathbb R^m$ is a ball, then we denote by $r(B)$ the radius of $B$ and $B^*$ is the ball with the same center as $B$ but with 5 times the radius of $B$.
Definition: A collection $\mathfrak B$ of balls in $\mathbb R^m$ is called Vitali cover of a set $E \subset \mathbb R^m$ if $E \subset \bigcup_{B \in \mathfrak B} B$ and for each $\epsilon > 0$ and $x \in E$ there is a $B \in \mathfrak B$ with $x \in B$ and $r(B) < \epsilon$.
So we now have a Version of the Vitali covering theorem, which says the following:
Theorem (Vitali covering theorem): Let $E \subset \mathbb R^m$ be a bounded subset and $\mathfrak B$ a Vitali cover of $E$. Then there exists a (possibly finite) sequence $(B_k)_{k \geq 1}$ of balls in $\mathfrak B$ which satisfies the following conditions:
(1) For $i \neq j$ we have $B_i \cap B_j = \emptyset$.
(2) $E \subset \bigcup_{k=1}^\infty B_k^*$
(3) $\lambda \left( E \big\backslash \bigcup_{k=1}^\infty B_k \right) = 0$, where $\lambda$ denotes the Lebesgue measure on $\mathbb R^m$.
I've worked through the proof of this theorem in the Real Analysis book by Makarov. As a remark after the proof he states that we can drop the requirement that $E$ is bounded. But how exactly do I see this? I've tried to reduce the statements to the bounded case, by looking at sets of the form $E_n := E \cap B(0, n) \backslash B(0, n-1)$, so that the Vitali covering theorem is true for these sets separately. But then, when $(B_k)_k$ is a desired sequence for $E_n$ and $(C_k)_k$ a desired sequence for $E_{n+1}$, we might have $B_i \cap C_j \neq \emptyset$, right? So have exactly can I generalize the theorem for unbounded sets?