Wikipedia proof of Vitali covering lemma uses Zorn's lemma. I was able to replace part of the argument which uses Zorn's lemma with another argument(I′ll provide the idea without the details in the end) which uses dependent choice only (for the case of $\mathbb{R}$ with the standard metric). Is it possible to prove the lemma for arbitrary separable metric space using some weaker version of choice (e.g. dependent choice) ? If not, is it possible to do this when the metric space is $\mathbb{R}^n$ with the standard Euclidean metric ? Unfortunately, I don't see any way to generalize my argument even to the $\mathbb{R}^n$ case.
The idea of the argument: Let′s show that any collection of nondegenerate segments with a lower bound $L>0$ on their length has a maximal subcollection of disjoint segments. Choose any segment $I_{0}$ from the collection. Because of the fact that there is a lower bound on the length of the segments we can choose a segment $I_{1}$(which is disjoint with $I_{0}$) to the right of the segment $I_{0}$ such that there is no segment between them which is pairwise disjoint with both of them. Continuing in the same manner we can construct at most countable sequence of segments. And after that we can do the same thing to the left of $I_{0}$. The union of these two sequences will give us the subcollection we need.
Vitali covering lemma and choice
axiom-of-choicemeasure-theorymetric-spacesseparable-spaces
Best Answer
You can generalize this to separable metric spaces. Take a dense sequence $(x_n)_{n∈ℕ}$ in your separable space (although I don't know if that's how you define separable in this setting). For each $x_n$, select a ball of your collection that contains it and that is disjoint from the ones already selected if it exists. Otherwise, select nothing. You end up with a maximal subset of disjoint balls.