Chavel's Riemannian Geometry: A Modern Introduction gives the following proof that the cut locus of a point $p$ in a Riemannian manifold $M$ has zero measure (Proposition III.3.1):
Indeed, the function $c(ξ)$ is continuous on all of $\mathsf{S}M$, so its restriction to $\mathsf{S}_p$ is certainly continuous. Thus, the tangential cut locus of $p$ is the image of the continuous map $ξ \mapsto c(ξ)ξ$ from $\mathsf{S}_p$ to $M_p$, and therefore has Lebesgue measure equal to 0. The image of the tangential cut locus of $p$ under the differentiable exponential map is the cut locus of $p$ in $M$, $C(p)$. Therefore, [...] for any $p \in M$, the cut locus $C(p)$ of $p$ is a set of measure zero.
Here, $\mathsf{S}M$ is the set of unit tangent vectors in $TM$, and $\mathsf{S}_p$ is the unit sphere in $T_pM$. The proof that $c(\xi)$ is continuous (Theorem III.2.1) only gives that $c(\xi)$ is $C^0$, not any higher differentiability.
I don't understand why the fact $C(p)$ is a continuous image of $\mathsf{S}_p$ is enough to imply that $C(p)$ has measure zero. I don't think zero-measure is preserved by continuous maps - there are Jordan curves (i.e. continuous images of the set $[0,1]$) with positive measure, and there are even graphs of continuous functions on the interval $[0,1]$ with Hausdorff dimension 2. Why does Chavel's proof work, or is there an easy way to modify his proof so that it works?
(I'm aware that Itoh and Tanaka prove 'The Lipschitz continuity of the distance function to the cut locus', i.e. that $c(\xi)$ is Lipschitz continuous, but I'd prefer a proof that doesn't depend on that. I want to use a similar approach to bound the dimension of a bisector $\{q \in M :dist(p_1,q)=dist(p_2,q)\}$.)