Approximation of Jordan curves Let $J$ be a Jordan curve in the plane, looked  upon as the homeomorphic image of the unit circle T. Suppose that for some  $\epsilon>0$ there is another
Jordan curve $K$ such that $K\subseteq V:= \{z: dist(z, J)< \epsilon\}$,
the $\epsilon$-neighborhood of $J$.
Do there exist parametrizations $j:T\to J$ of $J$ and $k:T\to K$ of $K$ such that  $|j(\xi)-k(\xi)|< C \epsilon$ for every $\xi\in T$?
 A: No. 
Actually I think you sort of didn't ask the question you meant to ask; we give a trivial counterexample to the question as asked and then a not quite as trivial counterexample to the question I think you meant to ask.
First, let $J$ be the unit circle, and let $K$ be the circle of radius $\epsilon/2$ centered at the point $1$. Every point of $K$ is close to a point of $J$ but the conclusion clearly fails.
What seems like the right version of the question assumes that every point of $K$ is close to a point of $J$, and that every point of $J$ is close to a point of $K$. Counterexample to that version:
Let $J$ be the unit circle.
Choose a very small number $\delta>0$. Let $r=1-\delta$, $R=1+\delta$. Define $K$ like so: Start at $Re^{i\delta}$. Run along the circle $|z|=R$ counterclockwise to the point $R=Re^{2\pi i}$. Take a straight line from $R$ to $r$. Run along the circle $|z|=r$ clockwise to the point $re^{i\delta}$, and then take a straight line back to $Re^{i\delta}$.

Edit We may as well parametrize $J$ by $g(e^{it})=e^{it}$. So we need to show there is no homeomorphism $f:S^1\to K$ with $|f(e^{it})-e^{it}|<\delta$. The best way to see this is to convince yourself that it's obvious; $e^{it}$ would have to be mapped to two different points of $K$. The following proof is imo far from the best way to see it, but it's by far the easiest thing to write down with complete rigor:
If $|f(e^{it})-e^{it}|=|f(e^{it})-g(e^{it})|<\delta<1$ then $f$ and $g$ are homotopic in $\Bbb C\setminus\{0\}$. So $f$ and $g$ have the same index, ie winding number, about the origin. Hence $0=1$.
