Assume $C$ is a Jordan curve in the plane, and let $S$ be a finite subset of its points.
Is there a simple proof that, for any $\epsilon>0$, there is a vector $\vec u$ of norm less than $\epsilon$ such that the shifted curve $C+\vec u$ does not intersect $S$ ?
Let $D=D(0,\epsilon)$ be the open disk of center 0 and radius $\epsilon$. Let $S=\{s_{1},\ldots,s_{n}\}$. Consider the first point $s_{1}$. Moving $C$ is equivalent to moving $s_{1}$ while letting $C$ fixed. Since $C$ is closed and of empty interior (see e.g. here), there is a nonempty open disk $U_{1}\subset D$ such that for $\vec u\in U_{1}$, $C+\vec u$ does not contain $s_{1}$. We can repeat this argument with the second point $s_{2}$ showing that there is a nonempty open disk $U_{2}\subset U_{1}$ such that, for $\vec u\in U_{2}$, $C+\vec u$ does not contain $s_{1}$ nor $s_{2}$. Continuing this way with the remaining points, we get the result.