Relative interior and dense subsets

Question

(This is a cross-post from here.) Let $A,B\subseteq \mathbb R^d$ be non-empty, such that $B\subseteq \overline A.$ For $S\subseteq\mathbb R^d$ define the relative interior of $S$ by $$\text{ri}(S)=\{s\in S\mid \exists\varepsilon>0:B_\varepsilon(s)\cap\operatorname{aff}(S)\subseteq S\}.$$ Here, $B_\varepsilon(x)$ is the open $d$-dimensional ball with center $x$, $\operatorname{aff}$ denotes the affine hull, $\operatorname{ri}$ denotes the relative interior and $\operatorname{conv}$ denotes the convex hull.

1. Do we have $\operatorname{conv}(A)\cap\operatorname{ri}(\text{conv}(B))\neq\{\emptyset\}?$
2. Or do we even have $A\cap \text{ri}(\text{conv}(B))\neq\{\emptyset\}?$

My intuition is the following (I failed proving it this way): We know that $\text{ri}(\text{conv}(B))$ is non-empty, hence let $x\in\text{ri}(\text{conv}(B)).$ Now $x$ can be approximated by elements in $\text{conv}(\overline A)$, hence also by elements in $\text{conv}(A).$ Choosing a convex combination $a$ close enough to $x$, this should be our solution.

Answer 1

As noted in the comments, the case where $A\subseteq\mathbb{R}^2$ is the interior of the unit square and $B$ is a side of that square satisfies the assumptions posed, but in that case neither (1) nor (2) holds. This example is easy to generalize, of course, and perhaps will lead you towards the claim you really want to make. Good luck.