Bounded convex sets $A,B\subseteq\Bbb R^n,n\ge 2$ with a common point, but disjoint boundaries

Question

Motivation of my question is this answer.

Let $A,B\subseteq\Bbb R^n,n\ge 2$ be bounded convex sets such that $\exists a\in A,$ and $a\in B,$ too and $\partial A\cap\partial B=\emptyset.$ Does it hold that $A\subset B$ or $B\subset A$?

I assumed:

1. $n\ge 2$ because, in $\Bbb R,$ segments $[0,2]$ and $[1,3]$ have disjoint boundaries, yet $2$ is an element of both. In $\Bbb R^2,$ however, their boundaries aren't disjoint.
2. $A,B\subseteq\Bbb R^2$ bounded because the stripes $A=\{(x,y)\in\Bbb R^2\mid 0\le x\le 2\}$ and $B=\{(x,y)\in\Bbb R^2\mid 1\le x\le 3\}$ is a counterexample.
3. $A,B\subseteq\Bbb R^2$ convex because $A=\{x\in\Bbb R^2\mid 0\le\|x\|\le 2\}$ and $B=\{x\in\Bbb R^2\mid 1\le\|x\|\le 3\}$ is a counterexample.

Now, if I'm not wrong, a convex set with more than $3$ non-collinear points (possible in $\Bbb R^n,n\ge 2$) should have a non-empty interior. Therefore one might consider two open balls of sufficiently smal radii (which are again convex) contained in $A$ and $B$ which the initial problem might boil down to, but I couldn't solve the problem that way.

I then tried to employ the boundedness of the boundaries (which with closedness make them compact) and try something with their distance, but to no avail.

Under what conditions does the statement hold and how can we prove it?

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Edit:

I'm not sure if this is relevant to the question, but according to this ResearchGate post boundary $\partial A$ of a bounded convex set $A\subseteq\Bbb R^n$ is homeomorphic to the unit sphere $\Bbb S^{n-1}$, hence connected (because it is the image of the connected set $\Bbb S^{n-1}$ under the continuous inverse of a map $f:\partial A\to\Bbb S^{n-1}, f(x)=\frac{x}{\|x\|}$?)

Answer 1

Our proof strategy will be as follows. Suppose that none of $A\setminus B$, $B\setminus A$, or $A\cap B$ is empty. We will exhibit a point in $B$ and on the boundary of $A$, and another point outside $B$ and on the boundary of $A$. We can then travel from the former point to the latter along the boundary of $A$ in a continuous manner. While we do so, we are bound to cross the boundary of $B$ somewhere. See the figure below.

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Formally: for the sake of contradiction, assume that $A\not\subseteq B$ and $B\not\subseteq A$. Since $A\cap B\neq\varnothing$, there exist $a\in A\cap B$ and $b\in B\setminus A$. Let $$\overline\lambda\equiv\sup\{\lambda\in[0,1]\,|\,(1-\lambda)a+\lambda b\in A\}.$$ I leave it as an exercise to show that $$x\equiv(1-\overline\lambda)a+\overline\lambda b\in \partial A\cap B.$$

Next, I claim that $\partial A\setminus B$ is not empty. If it is, then $\partial A\subseteq B$. Since every extreme point of $\overline A$ is also a boundary point (where $\overline A$ denotes the closure of $A$), the Krein–Milman theorem implies that $$A\subseteq\overline A=\overline{\operatorname{co}(\operatorname{extreme}(\overline A))}\subseteq\overline{\operatorname{co}(\partial(\overline A))}\subseteq\overline{\operatorname{co}(\partial A)}=\operatorname{co}(\partial A)\subseteq B,$$ contradicting $A\not\subseteq B$. Now take $y\in\partial A\setminus B$.

Since $\partial A$ is path connected for $n\geq 2$, one can find a continuous function $f:[0,1]\to\partial A$ such that $f(0)=x$ and $f(1)=y$. Let $$\overline c\equiv\sup\{c\in[0,1]\,|\,f(c)\in B\}.$$ Then, $z\equiv f(\overline c)\in\partial A\cap\partial B=\varnothing$, a contradiction.