Skeleton conjecture for higher-order Beer indices

For a set $X\subseteq\mathbb{R}^d$ and an integer $k\geq 0$, define

$$\operatorname{Skel}_k(X)=\bigcup_{Y\in\binom{X}{k+1}}\operatorname{Conv}(Y),$$

where $\binom{X}{k+1}$ is the set of $(k+1)$-element subsets of $X$. Let $\operatorname{b}_k(S)$ be the $k$-th Beer index of a set $S$, and let $\lambda_d$ denote $d$-dimensional Lebesgue measure. Skeleton conjecture. For every $k,d\in\mathbb{N}$ with $1\leq k\leq d$ and every $\varepsilon>0$, there is a $\delta>0$ such that if $S\subseteq\mathbb{R}^d$ satisfies $\operatorname{b}_k(S)\geq\varepsilon$, then there is a simplex $T$ with vertex set $X$ such that

$$\lambda_d(T)\geq\delta\lambda_d(S)\qquad\text{and}\qquad\operatorname{Skel}_k(X)\subseteq S.$$

The conjecture is known when $k=d$, since the $d$-skeleton is the whole simplex, and in the stated restricted one-dimensional simply connected case. Its validity in general remains open.