The argument of delta function comes in picture only when the electric field goes by 1/r^2.
let's consider a continuous charge distribution in space
when you are taking a continuous charge distribution,it will remain continuous to arbitrary precision. In the view of a mathematical construct, as a continuous function always remain continuous, No matter, by how much you magnify it, so is continuous distribution. So there is no sense of a single discrete particle in a continuous distribution. a continuous function is always continuous to arbitrary precision.
Now if we make a sphere and let the volume shrink arbitrarily to zero
when you are saying you are going to arbitrary zero, you are certainly not practical, so lets go mathematical,
mathematically, since there is no notion of single charged particle, no point of discreteness, the field not necessarily varies over 1/r^2. so Dirac delta is not going to Pop up.
By practical I mean , practical considerations of V-> 0
Practically, we take a lot more particle. You just add one more charged particle.
Now you have 2 charged particle, and just imagine it a dipole, now you will get a field varying over 1/r^3.Now also Dirac delta will not come into picture.one can see by calculation that even if one takes one barn by area then also it will contain more than one charged particle (say proton). So the argument of Dirac delta is shallow practically.
say, you are taking a linear charge distribution , when you are taking a volume element for your experiment , symmetrically your volume element itself should be something like line (like in Gauss law). Its no point in saying that I am taking a linear charge distribution and taking flux of deep down on electron, if you are concerned about one electron, then their is no point in making linear,spherical, areal distribution. If you using the properties of linear or planar distribution then you are not going to get delta function. you should consider at least thousand of electron so that you can atleast say
Oh! its seems planar, or linear
So, the flaw is You first said you took continuous charge (here,you are mathematical),
we calculate the divergence of point charge at origin
Then you took one charge (You are now practical),because mathematically, V-> 0 doesn't mean single charge. So It gave absurd result. Taking one charge is neither possible mathematically nor practically.**