Suppose $S$ is bounded star shaped domain of $ \mathbb R^n$ where $n \ge 2$ then $ \partial S$ is 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 $S$.
From my last question asked on MSE it turned out that boundary of a star shaped domain may not be Path connected.Now I am interested in knowing that: 

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

Edit: As a hint Someone suggested me to use the following exercise:
Let $X$ and $Y$ be topological space and $p: X \to Y$ be a quotient map and $p^{-1}(y)$ is connected for each $y \in Y$.Suppose $Y$ is connected then $X$ is connected.
But I am unable to see how to use this exercise here.
 A: Assume $x_0 = O$ (the origin). Using your hint, let $X = \partial S$ and $p : X\to \mathbb S^{n-1}$ be defined by $p(x) = \frac{x}{\|x\|}$. $p$ is obviously continuous. Also, $p$ is surjective as $S$ is an bounded open set. 
Now let $y \in \mathbb S^{n-1}$. Then the preimage $p^{-1}(y) $ consists of points of the form $t y$ for some $t>0$. Now let $0<t_1 <t_2$ so that $t_1y, t_2 y \in p^{-1}(y)$. We want to show that $ty \in p^{-1}(y)$ for all $t\in [t_1,t_2]$. 
To see that it suffices to show $ty \in \partial S$. But this is easy: As $t_2y \in \partial S$, there is $y_n \in S$ so that $y_n \to t_2 y$. As $S$ is star convex, $\frac{t}{t_2} y_n \in S$ for all $n$ and $0

$$\frac{t}{t_2} y_n \to \frac{t}{t_2} t_2y = ty$$
and hence $ty\in \partial S$ (Note that this shows in particular that $p^{-1}(y)$ is connected).
Lastly, we show:
Claim: $p$ is a quotient map. To see this, let $p^{-1} (A)$ be open set in $\partial S$. Then $p^{-1}(A) = W \cap \partial S$ for some open set $W \subset \mathbb R^n$. We can WLOG assume that $O \notin W$. Also we can assume that $W$ is a cone: $tW = W$ for all $t >0$. As $p : \mathbb R^n \setminus \{O\} \to \mathbb S^{n-1}$ is a quotient map, 
$$A = p(W\cap \partial S)= p(W)$$
and so $A$ is open. Thus $p$ is a quotient map. 
Using your hint, as $\mathbb S^{n-1}$ is connected for $n>1$, $\partial S$ is connected. 
A: Clearly you can assume that $x_0=0$. Suppose initially that $0\in int(S)$. 
Then $\partial S\subset \mathbb{R}^n\setminus\{0\}$, so $p:\partial S\to S^{n-1}$ defined by $p(x):=\frac{x}{|x|}$ makes sense and is continuous.
$p$ is a quotient map since it is closed (and this is because $\partial S$ is compact, so any closed set $C\subset \partial S$ is compact, so its image $p(C)$ is compact, hence closed).
Let $y\in S^{n-1}$. To show that $p^{-1}(y)$ is connected, we show that for any $\alpha y,\beta y\in\partial S$ the whole radial segment $[\alpha y,\beta y]$ connecting these two points belongs to $\partial S$. Clearly $[\alpha y,\beta y]\subseteq \overline{S}$ since $\overline{S}$ is star-shaped too. 
Now (assuming that $0<\alpha<\beta$) let $\gamma\in (\alpha,\beta)$.
If $\gamma y\in int(S)$, say $B_r(\gamma y)\subseteq S$, then applying the homothety $x\mapsto \frac{\alpha}{\gamma}x$ (which maps $S\to S$) we have $B_{\frac{\alpha}{\gamma}r}(\alpha y)\subset S$, i.e. $\alpha y\in int(S)$, contradiction. Thus $\gamma y\in \partial S$, so we have proved that $[\alpha y,\beta y]\subseteq \partial S$.
So from your exercise we know that $\partial S$ is connected. Now we remove the hypothesis that $0\in int(S)$: if $0\in\partial S$ choose a sequence $\epsilon_n\to 0$ and
put $S_n:=S\cup B_{\epsilon_n}(0)$. Then $S_n$ is star-shaped and $0\in int(S_n)$, so $\partial S_n$ is connected. Since $\partial S_n\setminus \overline{B_{\epsilon_n}}=\partial S\setminus\overline{B_{\epsilon_n}}$, you can easily deduce that $\partial S_n\to\partial S$ in the Hausdorff metric, so $\partial S$ is still connected (that connectedness is preserved passing to the limit is an easy exercise that you can solve even if you have no familiarity with Hausdorff distance).
Edit: maybe it's better to prove also that $\partial S_n\to\partial S$, for completeness. Obviously if $x\in \partial S_n\setminus \overline{B_{\epsilon_n}}$ we have $d(x,\partial S)=0$ (as $x\in\partial S)$, while if $x\in \partial S_n\cap \overline{B_{\epsilon_n}}$ then $d(x,\partial S)\le\epsilon_n$ (as $0\in\partial S)$. Thus $\max_{x\in\partial S_n}d(x,\partial S)\le\epsilon_n$. 
Analogously, if $x\in \partial S\setminus\overline{B_{\epsilon_n}}$ we have $d(x,\partial S_n)=0$. If instead $x\in \partial S\cap\overline{B_{\epsilon_n}}$, we can find some $y\in \partial S_n\cap\partial B_{\epsilon_n}$: this is because
$$\partial S_n=(\overline S\cup\overline{B_{\epsilon_n}})\setminus int(S\cup B_{\epsilon_n})\supseteq \overline{B_{\epsilon_n}}\setminus(S\cup B_{\epsilon_n})=\partial B_{\epsilon_n}\setminus S$$
and $\partial B_{\epsilon_n}\setminus S\neq\emptyset$, otherwise we would have $\partial B_{\epsilon_n}\subseteq S$, so by star-shapedness $\overline{B_{\epsilon_n}}\subseteq S$, which contradicts the assumption $0\not\in int(S)$. So we can pick any $y\in\partial S_n\cap\partial B_{\epsilon_n}$ and conclude that $d(x,\partial S_n)\le d(x,y)\le 2\epsilon_n$.
Thus $\max_{x\in\partial S}d(x,\partial S_n)\le 2\epsilon_n$. So finally
$$ d_H(\partial S_n,\partial S)\le 2\epsilon_n\to 0. $$
