Doesn't the fact that elementary particles are not black holes prove they are not point structures? De Schwarzschild radius of a mass $m$ is defined as  
$$r_s=\frac{2mG}{c^2}(m).$$
So if we insert in this formula the mass of an electron (a point particle, according to mainstream physics), which is $9.10*10^{-31}(kg)$, we get for $r_{sel}$, the Schwarzschild radius for the electron, (I've already done the calculation),
$$r_{sel}=13,7*10^{-58}(m).$$
Now let's look at the Planck length $l_p=1,62*10^{-35}(m)$.
How many times fits $r_{sel}$ in $l_p$? For the answer, we, of course, have to divide $l_p$ by $r_{sel}$. This gives:
$$\frac{1,62*10^{-35}}{13,7*10^{-58}}=1,18*10^{22}$$ 
If we multiply the electron's mass with this number we arrive at half of the Planck mass $m_p=2,18*10^{-8}(kg)$, so at $1,09*10^{-8}(kg)$.
On the other hand, if we put $m_p$ in the formula for the Schwarzschild radius we arrive at $3,23*10^{-35}(m)$, which is two times the Planck length.
The same results follow if we take the mass of electron neutrinos. 
Now the crux. If we don't measure anything (when we do that the Compton wavelength puts a limit on the accurateness with which we measure the position of electrons or whatever kind of point particle, and thus on the size of the electron, which is actually zero, so let's not measure anything and try to see without interfering what's going on; in our minds), doesn't the fact that the electron (or any other kind of point particle) has a Schwarzschild radius means that it is a black hole (like the other pointlike particles) and as such can't send out real photons so the real form of the point particles is not pointlike at all. It has to be greater than the corresponding $r_{sel}$. If this radius is $x$ times as big though $x$ times as fewer electrons fit in the Planck length $l_p$, so maybe it can be that the radius (or diameter) of an elementary point particle is equal to the Planck length and that the particles are such of form that they can, like circles on a cylinder (but some dimensions higher) can be put together without changing the form (two circles put on each other make a circle again, so if we put $1,18*10^{22}$ electrons together we have a particle with the Planck mass and the Planck length). 
EDIT
I made a mistake concerning the event horizon of charged black holes, so I want to correct that. Charged black holes (by which I mean black holes completely made up out of charged elementary particles like electrons or quarks; the last maybe can form a charged black hole as the color force between them is bigger than the electronic repulsion but let's not consider that case) can't exist.It would require electrons (or positrons) to gravitationally collapse until the mass is such that the electrons (which have a minimum mass with a maximum charge) form a black hole. The repulsion between the electrons is stronger than the gravitational attraction. So describing a non-existent electrically charged (let's consider such a hole built up entirely out of electrons) hole by the Kerr-Newman is actually a fantasy but let's take it for granted. And even if they did exist the potential energy of the electrons in a point would make the mass of the hole bigger, meaning that the event horizon gets bigger (this was the mistake I made because I wrote it became smaller). In fact, it would send the event horizon to infinity. And already when a black hole consists of two electrons, their infinite potential energy would send the event horizon to infinity.
The electrons in the (fantasy) charged black hole are considered in the SM as point particles (which I try to falsify). But how can the charge of a particle be put in infinite small point? Which brings us to the single electron. You would think that electric potential energy is infinite, but we are not bringing negative charge together to form the electron (which would explode under influence of the charge), giving rise to an infinite self-energy (or potential energy), which is the case when bringing together point charges in the hypothetical charged black hole. The electron isn't formed by bringing continuous charge together. But the charge can't be concentrated at a point. The electron charge doesn't contribute to its mass. So we can apply the formula for the Schwarzschild radius. Maybe we can even conclude that the electron's radius is equal to the Planck-length (like all other elementary particles).
For example, the charge can be distributed on an n-ball (with the radius equal to the Planck length) which can be put on each other like the circles (1-ball) on a cylinder can be put on top of each other without a distance between them, though the mean value of the distance between two circles on top of each other is half the diameter, so the Planck length. And maybe this is the reason that no distances smaller than the Planck length can be measured (which isn't to say smaller distances don't exist). On top of that, nobody knows for sure (nobody has probed those scales even far remotely, though there is a plethora of theories trying to quantize general relativity, without much success, I may add) if space-time fluctuations exist/become important at the Planck scales. It could be that space consists of hidden variables and as such gives rise to quantum mechanics (QFT), like the once strange erratic, and unpredictable and unexplainable behavior of the Brown's particle was eventually explained by the underlying motion of atoms and molecules. If that would be the case than space-time is not subjected to quantum mechanics because it is the cause of quantum mechanics.
 A: The Schwarzschild radius doesn't apply here, as the electron is not an uncharged singularity with no angular momentum. It has both charge and spin, so it is described by the Kerr-Newman metric instead. The Kerr-Newman metric has a minimum mass for which an event horizon is generated, and the electron's mass is far below this minimum value. As such, even if GR applies on this scale (and there's currently no guarantee that it does), the electron would not be a black hole.
A: Can you give a definition of a point particle? For example, a neutral atom has some size, right? What about electron? Can you specify its "size" from some scattering experiment? I would say, if exists an elastic cross section $\sigma$, one can speak of some size. In case of electron the elastic scattering is zero since the scattering is accompanied with some (at least soft) radiation. The inclusive cross section is approximately equal to the Rutherford cross section, but the "target" - electron is always broken in "pieces" - soft photons and in the relativistic case, soft and hard photons and neutral pairs. So it is never "point-like", but "compound". It is "big" and "soft". Note, this happens well before reaching the Plank distances.
