Here is a counterexample. Consider the closed 1st quadrant $Q = \{(x,y) \in \mathbb R^2 \mid x,y \ge 0\}$. Let $I_x = [0,1] \times 0$, and let $I_y = 0 \times [0,1]$.
I'm going to describe a Jordan arc $J$ with endpoints $(0,1)$ and $(1,0)$ such that $J \cup I_x \cup I_y$ is a Jordan curve bounding a compact set $X$, hence $X$ is homeomorphic to $\mathbb D^2$, and $(0,0)$ is a star point for $X$, but no interior point is a star point for $X$.
The arc $J$ is the polar coordinate graph of a function
$$r = f(\theta), \quad 0 \le \theta \le \pi/2
$$
with $f(0)=f(\pi/2)=1$. There will be an infinite, strictly increasing sequence
$$0 = \theta_0 < \theta_1 < \theta_2 < \cdots
$$
with $\lim_{i \to +\infty} \theta_i = \pi/2$, and I'll denote $a_i$ to be the point in the plane represented by polar coordinates $(\theta_i,r(\theta_i))$. The portion of $J$ with $\theta_{i-1} \le \theta \le \theta_i$ will be denoted $J_i$, and it is simply the Euclidean line segment
$$J_i = [a_{i-1},a_i]
$$
Thus, $J$ is obtained by adding the point $(0,1)$ to the concatenation of the closed line segments
$$J_0 J_1 J_2 \cdots
$$
As $n \to +\infty$ the segments $J_n$ will decrease in length to zero and approach $(1,0)$.
The idea is to construct the $J_n$'s so that the limit of the Euclidean slope of $J_n$ is equal to $+\infty$, but this limit is acheived in a "zig-zaggy" way that prevents any interior point of $X$ from being a center point.
Here's a few details on how to achieve this result. Consider $n \ge 1$. If $n$ is odd we require that $|a_{n-1}| < |a_n|$, and hence $J_n$ has "positive polar slope", because $r$ is increasing as a function of $\theta$ along the segment $J_n$. Also, if $n$ is even we require that $|a_{n-1}| > |a_n|$, hence $J_n$ has "negative polar slope". Finally, as $n \to +\infty$ we require that the Euclidean slope of $J_n$ approaches $+\infty$; hence for odd $n$ the polar slope is approaching $+\infty$ and for even $n$ the polar slope is approaching $-\infty$.
From these conditions, for any interior point $p \in Q$ one sees that for sufficiently large odd values, the point $p$ lies on the "outside" of the segment $J_n$, hence the line segment from $p$ to any point in the interior of $J_n$ does not lie in $Q$.
