It is answered a couple times as I can see but I want to know if my proof is working. I think I didn't understand the definition of trivial fundamental group very well.
Here it goes: A is simply connected if it is path connected and $\pi_1(X,a_0)=0$.
By the definition of star convex, there exists a line from $a_0$ to all $ x \in A$. Let $b,c \in A$ and let $f,g:I\to A$ st $f(0)=b$ $f(1)=a_0$ $g(0)=c$ and $g(1)=a_0$. Then there exists a path between $b$ and $c$ as $f \star g^-$. (Star is concatenation and line is reverse path.) Then A is path connected.
For the fundamental group to be trivial, I need to show that every loop can be shrinked down to a point. Or all the loops are homotopy equivalent. Let $\phi(s), s\in [0,1]$ be an arbitrary loop, based at point r. By the definition of star convex, there exists a path $m$ from r to $a_0$. Then $[a^-]\star[\phi]\star[a]$ is a loop at $a_0$, which means it is homotopolocigcally equivalent to $\pi_1(X,a_0)$. Then it is simply connected. $\square$
Most of the proofs that I have come up to create a homotopy between $a_0$ and the loop from an arbitrary point point on A, so would it be a better proof if I said: Let $H(s,t)$ be a homotopy which is defined as $H(,0)=\phi(),(,1)=_0$. Does that mean we are creating a homotopy between the loop at r and the boring loop at $a_0$. Thank you.
A star convex set $A$ with star point $a_0$ is path-connected, since any two points $x,y\in A$ can be connected via a path through $a_0$. It is also contractible: The map $H:I\times A\to A$, $(t,x)\mapsto tx+(1-t)a_0$ is a homotopy between $Id_A$ and $const_{a_0}$. Since $A$ is contractible, $\pi_1(A)=\{*\}$ and thus $A$ is simply connected.