Let $(A_i: i \in I)$ be a family of closed sets contained in $[0,1]$ such that for all $i,j \in I$ there exists $k \in I$ for which $A_i \cap A_j=A_k$. Denote by $\mathrm{co}(X)$ the convex hull of $X$. Is it true that $$ \bigcap_{i \in I}\mathrm{co}(A_i)=\mathrm{co}\left(\bigcap_{i \in I}A_i\right)\,\,\,? $$
Ps. Here a related question. Here, under weaker conditions, it is already shown that the right hand side is always contained in the left hand side. I guess a positive answer also on the converse inclusion.
The converse inclusion holds for any family $(A_i: i \in I)$ of compact subsets of a (Hausdorff) locally convex topological vector space $V$ (over $\Bbb R$). Indeed, let $a\in V\setminus \mathrm{co}\left(\bigcap_{i \in I}A_i\right)$. By Hahn–Banach separation theorem there exists a continuous linear map $\lambda : V\to \Bbb R$ and $t \in\Bbb R$ such that $\lambda(a) < t < \lambda (b)$ for all $b\in \mathrm{co}\left(\bigcap_{i \in I}A_i\right)$. Therefore $U=\lambda^{-1}(t,\infty)$ is a convex open subset of $V$ containing $\bigcap_{i \in I}A_i$ but not containing $a$. By so-called generalized Shura-Bura’s lemma, there exists a finite subset $J$ of $I$ such that $\bigcap_{i \in J}A_i\subset U$ (indeed, otherwise $(A_i\setminus U: i \in I)$ is a centered family of compact sets with empty intersection). Since $\bigcap_{i \in J}A_i=A_j$ for some $j\in I$, we have $\mathrm{co} A_j\subset U\subset V\setminus\{a\}$.