Assume we have a smooth bounded domain $\Omega \subset \mathbb{R}^d$, and two points at the boundary $x, y$. I want to show that there exists an integer $n$ such that one can draw an affine path joining $x$ to $y$ in $n$ steps. By an affine path I mean a path which picks up a direction at every contact with the boundary, follows this direction until it hits the boundary again, and pick a new direction at this rebound, and so on...
So in a strictly convex set, the number $n$ is 1 for any couple of points such that $x \ne y$, two for the couple $(x,x)$ (starting from $x$, we can rebound anywhere but at $x$ in one step and come back to $x$ for the second rebound). In general, starting from $x$, we can choose the direction, draw the corresponding segment inside the domain until a point $z \in \partial \Omega$, then choose another direction... and so on until we hit $y$. In general a result like the density of the set of orbits starting at a point will be a sufficient condition (but will also be way more than what we need, since at every rebound we can go from one orbit to the next one), but I am unsure that it holds without further conditions.
Also note that by connectedness, a path already exist, so it seems quite natural that an affine path should exist as well. I have tried to twist this path to obtain something but could not get what I was looking for.
You can think of pathological geometries such as the domain of $(x,y)$ defined by
$$0.9x^2\sin\frac\pi x\le y\le1.1x^2\sin\frac\pi x, x\in[0,1].$$
A path from $(0,0)$ to $(1,0)$ must make an infinity of bends, though the curve is everywhere differentiable.