Here's a stronger algebraic-topological argument. We can show that a star-shaped set $S$ is not just simply connected, it is contractible. There is an $\alpha$ so for each $p\in S$ the line segment $\gamma_p(t) = tp + (1-t)\alpha$ has $\gamma_p\in S$ for all $t$. Now define $H(p,t):S\times [0,1]\to S$ by $(p,t)\mapsto \gamma_p(t)$. This is a strong deformation retract of $S$ onto $\{\alpha\}$, so $S$ has the homotopy type of a point.
For every topological space $X$, every map $f:X\to S$ is homotopic to the constant map $X\to\{\alpha\}$. The intuition here is to just "slide" the map to $\alpha$, more precisely via the homotopy $F(x,t) = H(f(x),t)$. So in particular we can see that any map $\phi:\mathbb{S}^1\to S$ can be contracted to the constant map $\phi:\mathbb{S}^1\to \{\alpha\}$, hence $\pi_1S = 1$.
Here's a homotopy from $\gamma$ to a constant at $x_0$:
$$
H(x,t) = (1-t) \gamma(x) + t x_0
$$
At $t = 0$, this gives $\gamma$; at $t = 1$, ti gives a constant path at $x_0$.
Thus any two paths are homotopic (because both are homotopic to the constant path, by a homotopy analogous to $H$). Hence the start-shaped set is simply connected.
The only subtle points are
- Is the image of $H$ entirely within the set $S$? (Yes, by definition of star-shaped)
'
- For each fixed $t$, is $x \mapsto H(x, t)$ a smooth curve for $0 \le x \le 1$? (Yes, because its tangent curve is just $ x \mapsto t\gamma'(x)$, by the chain rule and product rule.
Added post-comments:
You've asked for a homotopy that preserves endpoints, so here goes. First, let's say that $x_1 = \gamma(0)$. Second, I'm going to use $t$ as the parameter for the path, and $s$ as the parameter for the homotopy, because that's what I'm used to.
Let
$$
H_0(t, s) = \begin{cases}
x_1 & 0 \le t \le \frac{s}{3} \\
\gamma(\frac{t-\frac{s}{3}}{1 - \frac{2s}{3}}) & \frac{s}{3} \le t \le 1 - \frac{s}{3} \\
x_1 & \frac{s}{3} \le t \le 1
\end{cases}
$$
Then $H_0$ is a homotopy from $\gamma$ to "$\gamma$-traversed-three-times-as-fast-with-a--pause-before-and-after-traversal" Let's call that new curve $\gamma_1$, so
$$
\gamma_1(t) = H_0(t, 1).
$$
Now let
$$
H_2(t, s) = \begin{cases}
(1-s) x_1 + s ((1-3t)x_1 + 3t x_0)) & 0 \le t \le \frac{1}{3} \\
(1-s) \gamma_1(t) + s x_0 & \frac{1}{3} \le t \le \frac{2}{3} \\
s x_1 + (1-s) ((3-3t)x_0 + (3t-2) x_1))& \frac{2}{3} \le t \le 1
\end{cases}
$$
That's a homotopy from $\gamma_1$ to a curve that goes from $x_1$ to $x_0$ in the first third, then sits at $x_0$, then returns to $x_1$ during the last third. Let's call that new curve $\gamma_2$.
Finally, let
$$
H_2(x, t) =
(1 - t) \gamma_2(x) + tx_1.
$$
That's a homotopy from $\gamma_2$ to the constant path at $x_1$.
We sequence these to get a homotopy from $\gamma$ to the constant path at $x_1$, and the homotopy, as you'll observe, is endpoint preserving, i.e., it's a homotopy of loops, not just paths.
Best Answer
Half of an annulus will do the job. In terms of an equation, $$\{(x, y) \in \Bbb R^2 : 1 \leqslant x^2 + y^2 \leqslant 2, \ x \geqslant 0\}.$$ Do you see why this satisfies both your conditions?