Expanding EM field of a charged particle

Would this not be a solution to the age old problem of the total energy of the coulomb field being infinite, since the fields do not actually permeate throughout an infinite space?

No, because the infinite space isn't the problem. If we try to integrate the field energy of a static charge over all space, we have $$ E = frac{epsilon_0}{2} iiint E^2 , dV = frac{epsilon_0}{2} left( frac{q}{4 pi epsilon_0} right)^2 iiint frac{1}{r^4} r^2 sin theta , dr , d theta , d phi = frac{q^2}{8 pi epsilon_0} int_0^infty frac{dr}{r^2} $$ If we split this last integral up into two pieces relative to some radius $R > 0$, we get $$ E = frac{q^2}{8 pi epsilon_0} left[ int_0^R frac{dr}{r^2} + int_R^infty frac{dr}{r^2} right] $$ The second integral is finite, no matter what $R$ is (it's equal to $1/R$); and the first integral is infinite, no matter what $R$ is. This means that the divergence of the total field energy is not due to the field's infinite volume, but rather due to its infinite density near the point charge.