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Tapio Rajala
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The Besicovitch covering theorem fails for example in the Heisenberg group, see [ E. Sawyer and R. L. Wheeden, Weighted inequalities for fractional integrals on Euclidean and homogeneous spaces, Amer. J. Math. 114 (1992), no. 4, 813–874. http://www.jstor.org/stable/2374799 ]

The following was more of a comment to Misha's question.

It is easier to see (without assuming the space to be a length-space) that the Lemma 4.2.2. in http://www.math.wustl.edu/~sk/books/root.pdf is not true for general complete doubling metric spaces nor is the Besicovitch covering theorem:

Take for instance a space $X = \mathbb{N} \cup\{0\}$ with the distance $$d(0,j) = 2^{-j} \text{ for } j \ne 0$$ and $$d(i,j) = 2^{-j}+2^{-i} \text{ for }0 \ne i\ne j \ne 0.$$

To see that this does not satisfy the Lemma nor the Besicovitch covering theorem consider for any $k \in \mathbb{N}$ the collection $\{B(j,2^{-j}+2^{-k}) ~:~ j = 1, \dots, k-1 \}$.

Tapio Rajala
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