Why does no object end up at the center of their circular paths?
If the path is circular then, by definition, the particle maintains a constant distance from the center of the path. Perhaps you're asking why the particle has a circular path?
Would someone have a rigorous proof that a constant centerwards force
produces a circular orbit
One needs something more for a circular orbit; one needs a force that is constant in magnitude and (always) perpendicular to the particle's velocity.
(The question of whether such an orbit is stable against perturbations is another question entirely.)
Perhaps the easiest way to think about this is to solve for the acceleration for a particle with uniform circular motion.
For example, in Cartesian coordinates, such circular motion is of the form
$$x(t) = R\cos(\omega t - \delta)$$
$$y(t) = R\sin(\omega t - \delta)$$
where $R,\;\omega$ and $\delta$ are constants Then, the acceleration components are
$$a_x(t) = -\omega^2 R\cos(\omega t - \delta)$$
$$a_y(t) = -\omega^2 R\sin(\omega t - \delta)$$
The magnitude of the acceleration is then
$$a = \omega^2 R$$
and it is clear that the acceleration is constant in magnitude.
The velocity components are
$$v_x = -\omega R \sin(\omega t - \delta)$$
$$v_y = \omega R \cos(\omega t - \delta)$$
and it's now easy to see that the acceleration is perpendicular to the velocity
$$\vec v \cdot \vec a = (\omega^3 R^2 - \omega^3 R^2)\sin(\omega t - \delta)\cos(\omega t - \delta)=0$$
We've shown that uniform circular motion implies a constant magnitude acceleration that is (always) perpendicular to the velocity.