Doesn't electron capture imply an electron is not a point charge?
It needs to have a radius that overlaps with the proton. If it was a point charge, no matter how close it got to the proton, the electron will never overlap with the proton, and electron capture would be impossible.
Your reasoning is wrong, but you are right to be cautious about the notion of a 'point charge'.
First of all, let's note that a proton is not a point charge. So a point-like thing could easily overlap with a proton.
Next let's note that, in quantum theory, the locations of things are described by wavefunctions and in practice the idealised perfectly narrow distribution (Dirac delta function) never occurs. It does not occur because it would require infinite energy. So in this sense electrons are not point-like because no material entity of any kind can be in a point-like position distribution.
On the other hand it is currently believed that electrons have no sub-structure and furthermore in ordinary quantum mechanics they are treated as point-like in a technical sense, in that one imagines the total set of processes in which they can in principle take part includes those infinitely narrow distributions. In other ideas for basic physics, such as string theory, the situation is a bit different.
Finally, in classical physics there are also difficulties with point-like objects, because they would have infinite gravitational and electrical energy.
So, in short, we don't expect truly point-like things anywhere in physics.
In my own opinion, you need to consider both descriptions : wave-like and point-like to fully understand what is going on at the atomic scale. There are several examples where we need a point-like charge to fully understand some phenomena. Here are two example where a translational and a rotational symmetry are involved.
In solid state physics, a strongly correlated system (where Bloch's translational symmetry is broken) is described by a coulomb potential acting a point-like particle hopping from one atomic site to another making a transition from a metal to and insulator (Hubbard Hamiltonian). A wave-like description cannot lead to such a simple and elegant description perfectly in agreement with experimental measurements.
The orbital magnetic moment and the orbital angular momentum can be described classically by a point-like particle rotating. Obviously, a wavy description is needed for the quantization of the angular momentum.
To summarize a point-like description is often essential to describe the motion leading to the stationary state and a wave-like description encompasses the symmetry (translational or rotational) of this motion in the wave function leading to the ground state.
It might not be the answer of your question by I hope it helps.
The notion "point particle" is coming from classical mechanics, where all fields $F(t,x)$ acting on a point particle, identified by its enumerated center of mass coordinate $x_k$ with momentum $p_k$, can simply expressed as $F(x_k)$ by replacing the field coordinate with the coordinate of the particle passing that point.
Gravitational and electric interactions can be formulated in terms of the kinetic momentum $p_k - \frac{e}{c}\ A(x_k) $ involving corrections to the canonical momentum $p$.
The central entity is the quadratic norm of of the kinetic momentum, aka "mass times transport velocity" $$(p(x_k)-A(x_k))^2 = G(x_k)((p(x_k)-A(x_k)),(p(x_k)-A(x_k)) )$$ as a simple square, whrere again only the center of mass coordinate $x_k$ determines the gravitational effects at a point in space.
This classical notion of "point particles with no internal structure" is transfered, by the principle of first quantizaton (Heisenberg, Schrödinger, Dirac), to quantum mechanics.
In quantum mechanics, paths of position and momentum in phase space are replaced by position operator $$X_k: \ f(x_1,\dots,x_n) \to \ x_k \ f(x_1,\dots,x_n)$$ and momentum operator $$P_k\ f(x_1,\dots,x_n)\to -i \ \partial_{x_k}\ f(x_1,\dots,x_n)$$
and to the astonishment of the public, again the external electric and graviatational fields are pointwise encoded in the square of the momentum operators
$$ G(x_k)(P_k - \frac{e}{c}\ A(x_k)), P_k - \frac{e}{c}\ A(x_k))$$
The states in quantum mechanics can be as diluted or concentrated in space as external field conditions deman, making obsolate all speculations about their the electric field energy of their own charge.
The first step in the interaction of many electrons in these wave functions of many variables, is to eliminate the electric self energy.
The reason for this step is clear: Quantums first priciple since Planck is the energy of photons with fixed frequency $$ \hbar \omega_k = E_k$$ But the Fourier representation of the static Coulomb field has $\omega_k=0$ for all euclidean modes $k$ with density $\frac{1}{k^2}$. They simply don't take part in the energy exchange game.
