TITLE: Proof that a set is non-star-shaped.
QUESTION [9 upvotes]: Note: A subset $S\subseteq \mathbb{R}^2$ is called star-shaped if there exists an element $x\in S$ such that for every $y\in S$, the line segment connecting $x$ to $y$ is contained in $S$.
We want to give an example of a non-star shaped subeset of $\mathbb{R}^2$ and carefully prove it is not star-shaped.

I have begun by giving my example of a set that is not star shaped. It is the set of all points between the two circles, $x^2+y^2=1$ and $x^2+y^2=4$.
This set looks like an Annulus, as shown below.

To me, it is obvious that this set is not star shaped, because there is no point in the set that will not have to cross the circle in the middle of the set in order to make a line segment to every other point in the set. 
I don't know how to prove this, however. Perhaps it is easier to prove giving a different example of a nonstar shaped set.

Any help would be greatly appreciated.

REPLY [17 votes]: Note that the figure $S$ is centered at $0$. To show that it is not star-shaped, note that $x\in S$ if and only if $-x\in S$. So the striaght line joining $x $ and $-x$ must pass the origin, which is not in $S$. So $S$ is not star-shaped. 
Heuristically, everything that has a hole in between will not be star-shaped. So, as Brian suggests, you can use also something like $S = \{x^2 + y^2 = 1\}$, and the proof is the same.