Suppose $AB$ and $BC$ are two successive chords of the ball's path. Then by the reflection law $\angle ABO = \angle OBC$. But $OAB$ and $OBC$ are isosceles and so $\angle AOB = \angle BOC$. Hence $AB = BC$. So every chord of the path is the same length $d$.

We now claim that through any given point $P$ inside the circle there are at most two chords length $d$. Let $AB$ and $CD$ be a chord containing $P$, with $AP = a$ and $CP = b$. The power of $P$ with respect to the circle is $PA \cdot PB = PC \cdot PD \iff a(d-a) = b(d-b) \iff a = b$ or $a+b = d$. This means that $P$ always divides the chords containing it in two segments of fixed lengths $a$ and $d-a$. Now if three chords passes through $P$, the circle with center $P$ and radius $a$ would cut the circle of the billiard table three times, a contradiction.

Thus if the path passes through $P$ more than twice, then on two occasions it must be moving along the same chord $AB$. That implies that $\angle AOB$ is a rational multiple of $2\pi$ and hence the path will traverse $AB$ repeatedly.