Hausdorff measure is countably additive on every subset

By the Vitali theorem there is a nonmeasurable $V$. Therefore some $T\subseteq \mathbb{R}$ fails its Carathéodory splitting identity. Let $A=T\cap V$ and $B=T\setminus V$; these are disjoint and their union is $T$. Thus Lebesgue outer measure fails finite additivity on this pair. Full Choice here supplies in particular the Countable Choice hypothesis of the line comparison.

The line equality transfers this failure to ${\mathcal{H}}^1$. Add empty sets after $A,B$ to make a disjoint sequence; its sum is still ${\mathcal{H}}^1(A)+{\mathcal{H}}^1(B)$ and differs from ${\mathcal{H}}^1(T)$. Neither piece can be empty, since such a splitting would be automatic. Hence countable additivity on all subsets is false.