I'm reading a proof given by Edward Szpilrajn found here and it says, among other things, that if $A$ is a non-empty set, and if $m\in\mathbb{N}$ are such that $$(*)\quad \mathcal{H}^{m+1}(A) = 0 ,$$ then we have $$(**)\quad \mathcal{H}^m(S_r) = 0 $$ for almost all $r>0$, with $S_r = \{x\in A: |x-x_0|=r\}$. Here, $\mathcal{H}^h$ denotes the $h$-dimensional Hausdorff measure.
Supposing it's true, the author claims that, by induction, it shows that $\dim(A) \leq m$, where $\dim(A)$ denotes the topological dimension of $A$.
Can you help me to understand how the induction is sufficient to prove the claim?