Using the general ham sandwich theorem to proof Hobby-Rice

Question

Matousek mentions that you can proof the continuous necklace theorem known as Hobby-Rice theorem via the continuous ham sandwich theorem. The continuous ham sandwich states:
Let $\mu_1,\mu_2,...,\mu_d$ be finite Borel measures on $\mathbb{R}^d$ such that every hyperplane has measure 0 for each of the $\mu_i$. Then there exists a hyperplane h, such that $$\mu_i(h^+) = \frac{1}{2}\mu_i(\mathbb{R}^d)$$ for i=1,...,d where $h^+$denotes one of the half-spaces defined by $h$.
The Hobby-Rice theorem states:
Let $\mu_1,\mu_2,...,\mu_d$ be continuous probability measures on $[0,1]$. Then there exists a partition of $[0,1]$ into $d+1$ intervals $I_0,I_1,...,I_d$ and signs $\varepsilon_0,\varepsilon_1,...,\varepsilon_d \in \{-1,+1\}$ with $$\sum_{j=1}^{d} \varepsilon_j \cdot \mu_i(I_j)=0$$ for $i=1,2,...,d$

Does anyone know a simple proof for this? Thanks.