Line segments intersecting Jordan curve I have thought this problem a week without success.
Is there a set $A\subset \mathbb{R}^2$ such that 


*

*The boundary of $A$, $\partial A$, is a Jordan curve and

*For any $B\in \operatorname{int} A\ne\emptyset $, $C\in \operatorname{ext} A\ne\emptyset$ , the line segment $BC$ intersects $\partial A$ infinitely many times?


Any ideas?
 A: This is not an answer, only some ideas. 
Let $A$ be a Jordan domain, and let $f\colon \mathbb D\to A$ be a conformal map of the unit disk onto $A$. By Caratheodory's theorem $f$ extends to a homeomorphism of closures. One says that $f$ is twisting at a point $\zeta\in\partial\mathbb D$ if for every curve $\Gamma\subset \mathbb D$ ending at $\zeta$ the following holds: 
$$\liminf_{z\to\zeta}\ \arg(f(z)-f(\zeta))=-\infty, \quad
\limsup_{z\to\zeta}\ \arg(f(z)-f(\zeta))=+\infty,\quad (z\in\Gamma) $$
(This definition is from the book Boundary behaviour of conformal maps by Pommerenke. Some sources have different wording. The definition is unchanged if one replaces $\Gamma$ by a radial segment.)
If $f$ is twisting at $\zeta$, there is no line segment that crosses $\partial A$ only at $f(\zeta)$. Indeed, the preimage of such a line segment would be a curve along which $\arg(f(z)-f(\zeta))$ is constant. 
Pommerenke's book presents several results on twisting points and gives pointers to literature. The message is that $f$ can have a lot of twisting points. Of course, it is impossible for $f$ to be twisting at every point of $\partial\mathbb D$. (Consider any disk contained in $A$ whose boundary touches $\partial A$.) But what we want is for every point of $\partial A$ to be twisting either on the inside or on the outside (i.e., for the conformal map onto interior or onto exterior). 
My profound lack of knowledge of complex dynamics suggests that the Julia set of the quadratic polynomial $p(z)=z^2+\lambda z$ could have this property when $|\lambda|<1$ and $\mathrm{Im}\,\lambda\ne 0$. Indeed, in this case the polynomial has two Fatou components, and the boundary between them is a Jordan curve, indeed a quasicircle. The curve appears to be twisting as expected (see this applet, which takes polynomials in the form $z^2+c$. Here $c=\lambda/2-\lambda^2/4$ with $|\lambda|<1$, so we are in the main cardioid of the Mandelbrot set). Maybe someone who understands complex dynamics can tell if this Julia set is indeed an example.
A: I am not sure, but here's an idea. Take the graph of $x sin(1/x)$ and join the two zeroes around (+/-)0.15 by a big enough curve. Take the points to be $(0.12, 0)$ and $(-100, 0)$.
EDIT: Misunderstood, that it is required for only a single pair of points. OP wants this to happen for every pair of points.
A: Such a set does exist. See https://mathoverflow.net/questions/100025/how-many-times-line-segments-can-intersect-a-jordan-curve
