In a metric space $(M,d)$, I have compact nonempty sets $A, B$ and $C$ with $A\subset \operatorname{int} B$ and $B \subset \operatorname{int} C$, where $\operatorname{int}$ denotes the interior of a set and $\partial$ its boundary. I'm trying to find out if $d(A,C - \operatorname{int} B) = d(A, \partial B).$ If this were to be true it would solve a problem I'm working on, but can't prove it nor find a counterexample. It is always true that $d(A,C - \operatorname{int} B) \leq d(A, \partial B),$ since $\partial B \subseteq C- \operatorname{int} B, $ and the reverse inequality seems true because taking a point in $C - \operatorname{int} B$ which is not in the boundary of $B$, it would be "further away" from $A$ than a point in the boundary of $B,$ but I'm thinking geometrically on the plane, and I can't work on the triangle inequality to give me this result.
Is what I'm trying prove to even true? Any hints are appreciated, thanks.
(OBS: the metric space is also a smooth manifold in my problem, if any extra structure helps).
(OBS2: in the original question, $M$ could be a topological manifold, since I thought this problem could be solved with topology alone. User @theo-bendit showed this is false in general, but more research made me see this may be true for a smooth manifold, as in this other question When is distance to the boundary always less than that to the exterior? . In not used to work with riemann metrics so I can't be sure).
I can't comment on the topological manifold part here, but in general, no, this is not the case.
Let $K$ be the Cantor Middle Third set, and \begin{align*} M &= C = K \cup [100, 103] \\ B &= \left(K \cap \left[0, \frac13\right]\right) \cup [101,102] \\ A &= K \cap \left[0, \frac19\right]. \end{align*} Then, \begin{align*} C \setminus \operatorname{int} B &= \left(K \cap \left[\frac23, 1\right]\right) \cup [100, 101] \cup [102, 103] \\ \partial B &= \{ 101, 102\} \\ d(A, C \setminus \operatorname{int} B) &= \frac59 \\ d(A, \partial B) &= \frac{908}{9}. \end{align*}