I take him to mean that starting out the derivation of the expressions for electrostatic energy with the expression $\frac{q_1q_2}{4\pi \epsilon_0 r_{12}}$, i.e. electrostatic energy equals the work required to bring two charges together., from this and the principle of superposition, you can get $U = \frac{1}{2}\int \rho \,\phi \,dV$, where $\phi$ at a point is made up of the contributions from all other charges than the charge $\rho \, dV$ at that point. It could have ended there with this expression (with energy maybe located where the charges are, i.e. $\phi \, \rho$) and point charges would not give us any headache. But because of the additional fact that we also have that $U = \frac{\epsilon_0}{2}\int E \cdot E \,dV$, and that we can interpret this as there being an energy density $\frac{\epsilon_0}{2}E^2$ associated with the field, and because of the fact that this expression is actually the correct distribution of enery (which is consistent with, but cannot be deduced from electrostatics), because of this it follows that even a single point charge, with no other charges around, will also have energy in its field, namely with density $\frac{q^2}{32\pi^2 \epsilon_0 r^4} $ at a distance r (and the integral of this diverges).

I take him to mean that starting out the derivation of the expressions for electrostatic energy with the expression $\frac{q_1q_2}{4\pi \epsilon_0 r_{12}}$, i.e. electrostatic energy equals the work required to bring $two$ charges together, from this and the principle of superposition, you can get $U = \frac{1}{2}\int \rho \,\phi \,dV$, where $\phi$ at a point is made up of the contributions from all other charges than the charge $\rho \, dV$ at that point. (So far in his derivation, even in this expression, there is only energy $between$ charges, and an isolated point charge in the integral therefore does not make sense). It could have ended there with this expression (with energy maybe located where the charges are, i.e. $\phi \, \rho$) and point charges would not give us any headache. But because of the additional fact that we also have that (numericaly at the least) $U = \frac{\epsilon_0}{2}\int E \cdot E \,dV$, and that we can interpret this as there being an energy density $\frac{\epsilon_0}{2}E^2$ associated with the field, and because of the fact that this expression is $actually$ the correct distribution of enery (which is consistent with, but cannot be deduced from electrostatics), because of this it follows that even a single point charge, with no other charges around, will also have energy in its field (so the energy isnt only $between$ charges anymore), namely with density $\frac{q^2}{32\pi^2 \epsilon_0 r^4} $ at a distance r (and the integral of this diverges).