Q:

Suppose $S$ is bounded star shaped subset of $ \mathbb R^n$ where $n \ge 2$ then $ \partial S$ is path connected

Def: A set $S$ in the Euclidean space $ \mathbb R^n $ is called a star shaped domain if there exists $x_o$ in $S$ such that for all $x$ in $S$ the line segment from $x_0$ to $x$ is in $S$.
Some days ago in a comment in the answer to one of my questions @Hagen Von Eitzen commented that  

Suppose $S$ is bounded  star shaped subset of $ \mathbb R^n$ where $n \ge 2$ then $ \partial S$ is Path connected.

I am unable to prove this result. Please help!

A:

Here's another example. Let $(r,\phi)$ be the polar coordinates in $\mathbb R^2$. Define $S$ as the closure of $$\left\{(r,\phi)\mid r\leq2+\sin\frac{30}{\phi(2\pi-\phi)}\right\},$$ where we only allow $r\geq 0$ and $\phi\in(0,2\pi)$. See here for a picture.
This is star shaped pretty much by definition (since for each $\phi$ it contains all the points of at most a certain distance from the origin) and its boundary is a variant of the Warsaw circle and is not path connected.

A:

It seems that this is not true. 
For example, consider the following subset in $\mathbb R^2$. 
For each $n\in \mathbb N$ consider the points $p_n = (\frac 1n, 1)$ and $p_{n+1}=(\frac{1}{n+1}, 1)$. Then consider the closed polygon $U_n$ formed by the vertices 
$$O=(0,0), p_{n+1}, q_n = \left(\frac{2n+1}{4n(n+1)}, \frac 12\right), p_n$$
Note that this is a star convex set, with the center $(0, 0)$. (The third point is chosen this way: Let $D$ be the midpoint of $p_n, p_{n+1}$. Then the line $OD$ will hit $y=\frac{1}{2}$ at one point, which is $q_n $). 
Now let 
$$U_1 = \overline{\bigcup_n U_n} = \bigcup_n U_n \cup \{(0,y): 0\le y\le 1\},$$  and let 
$$U = \{(x, y): (|x|, |y|)\in U_1\}\cup\{(x, y): |y|\le |x|\le 1\}$$ 
Then $U$ is a closed star convex set with center $(0,0)$, but the boundary is not path connected (only connected). The reason is that the boundary contains the segment $\{(0, y): \frac 12\le |y|\le 1\}$ and the boundary on both sides converges to these two segments as in the case of topological sine curve. 

