Is a set "trapped" between two convex sets, convex?

As stated, this is false.

In $R^3$, let $C = \bar{B}(0,1)$ and let $D = C \cap Q^3$. Then $C$ is the closure of $D$ and the interior $A$ of $D$ is empty. There are many non-convex bodies $B$ contained in $C$.

Let $C$ be closed and strictly convex with $int \ C \ne \emptyset$. Let $B$ be such that $$ int \ C \subset B \subset C. $$ The interior of a set is an open set, so $int\ C \subset int \ B$.

Then $B$ is strictly convex: let $x,y\in B \subset C$. Then $\lambda x + (1-\lambda)y \in int \ C\subset int \ B$ for $\lambda \in (0,1)$.