Prove that external charge has no net flux on a gaussian surface mathematically

Question

I understand how the field lines enter and leave the gaussian surface. But my concern is that the field isn't constant everywhere on the gaussian surface, i.e, there exactly doesn't exist an $- Ecdot da$ corresponding to every $Ecdot da$. I understand the idea of how the enlargement of the area compensates for the reduction in the field. But I want it mathematically.
If you could prove that for a point charge on an irregular surface, the job would be done.
I have however, made an attempt to prove it in the below image, where I basically proved how the flux doesn't depend on $r$ for an infinitesimally small surface $da$. But I am not satisfied with it. Please help. Ignore the proof if it isn't good enough and doesn't make sense, I am just a fresher, hence a noob.
[Proof]1

Urb · Answer

Suppose we have a point charge $q$ at the origin $vec{r}=0$. Then choose an arbitrary Gaussian surface $S$ enclosing a volume $V$. By definition of flux, the electric flux through the surface is
$$Phi=iint_Svec{E}cdotvec{dS}$$
By the divergence theorem, this is equal to
$$Phi=iiint_Vnablacdotvec{E} dVtag{1}$$
Then, since we know the form of $vec{E}$, namely
$$vec{E}=frac{1}{4pivarepsilon_0}frac{q}{r^2}hat{r}$$
we can calculate directly its divergence
$$nablacdotvec{E}=frac{q}{4pivarepsilon_0}nablacdotleft(frac{hat{r}}{r^2}right)=frac{q}{varepsilon_0}delta^3(vec{r})tag{2}$$
where in the last step I have used the mathematical identity1
$$nablacdotleft(frac{hat{r}}{r^2}right)=4pidelta^3(vec{r}).$$
Inserting $(2)$ in $(1)$ we have
$$Phi=iiint_Vfrac{q}{varepsilon_0} delta^3(vec{r}) dV$$
And finally, if the surface does not enclose the charge, i.e., $vec{r}=0notin V$, the last integral vanishes due to the translation property of the Dirac delta2.

1Take a look at this Math.SE post for details.
2Here it is $$iiint_{V}delta^3(vec{r}-vec{r}_0) dV=begin{cases}0quadtext{if }vec{r}_0notin V1quadtext{if }vec{r}_0in Vend{cases}$$

Prove that external charge has no net flux on a gaussian surface mathematically

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