Suppose that $(X,d)$ is a complete metric space and that $f \colon X \to X$ is an injective contraction. That is, there exists $K \in (0,1)$ such that $0 < d(f(x),f(y)) \le K d(x,y)$ for all $x,y \in X$. Let $S \subset X$ be a set with non-empty interior. Is is possible for $f(S)$ to have empty interior?
My topological skepticism leads me to suspect that there is probably a clever counterexample.
Let $X$ be the subset of $\mathbb R^2$ consisting of the unit square $[0,1] \times [0,1]$ and the line segment $[0,1] \times \{2\}$, and let $f(x,y) = (x/2,y/2)$. Then the line segment $[0,1] \times \{2\}$ has interior $(0,1) \times \{2\}$ in $X$, but its image $[0,1/2] \times \{1\}$ has empty interior, since every point $(x,1)$ in that image has points of the form $(x,1-\epsilon)$ arbitrarily close to it.