A bounded subset in $\mathbb R^2$ which is “nowhere convex”?

Let $F : \mathbb S^1 \to \mathbb R^2$ represents a simple closed curve $C$ in $\mathbb R^2$. The Jordan curve theorem says that the curves bounds a interior domain $\Omega$ and $\partial \Omega= C$. Do we always have two points $a, b\in C$ such that the line segment joining $a$ and $b$ lies completely in $\Omega$? What’s more, can we have

“For all $\epsilon>0$, there is two points $a ,b\in C$ such that $|a- b| < \epsilon$ and the line segment joining $a$ and $b$ lies completely in $\Omega$”.

It seems that it is related to how the $C$ oscillate. If $C$ is some kinds of fractal curves, it might happens that $C$ is nowhere convex. What if we assume that $F$ is of bounded variation?

This question is related to the following question:

Uniqueness of curve of minimal length in a closed $X\subset \mathbb R^2$

Solutions Collecting From Web of "A bounded subset in $\mathbb R^2$ which is “nowhere convex”?"

There is no “nowhere convex” subset, and bounded variation is not needed.

Claim. Let $\Omega \subset \mathbb R^n$ be a bounded open set. Then for every $\epsilon>0$ there exist two points $a,b\in \partial \Omega$ such that $|a-b|<\epsilon$ and $\{(1-t)a+tb:0<t<1\}\subset \Omega$.

Proof. Let $r = \sup \{|x| : x\in {\Omega}\}$. For a small $\epsilon>0$, pick a point $p\in \Omega$ with $|p|>r-\epsilon$. Let $L$ be a line through $p$ in a direction orthogonal to the line segment $[0,p]$. The intersection $L\cap \Omega$ is an open subset of $L$. Since $\Omega\subset B_r = \{x: |x|< r\}$, the Pythagorean theorem implies $$\operatorname{diam}(L\cap \Omega) \le \operatorname{diam}(L\cap B_r) = 2 \sqrt{r^2 – (r-\epsilon )^2}$$
which can be made arbitrarily small. Any connected component of $L\cap \Omega$ is a line segment contained in $\Omega$ and with endpoints $a,b\in \partial \Omega$. $\quad \Box$