I am trying to find a formal proof for the following property of Dirac delta function:
for any function $f$ : $$\int_{-\infty}^{+\infty} \delta(x)f(x)dx=f(0),$$ where $\delta$ is Dirac delta generalized function.
Sorry if the above $f$ must have some properties but I think it shouldn't. That is what I want the whole text and the proof.
Is there anyone who knows?
Indeed, as you were told, there is no proof of the formula you're asking, this is a definition. However, here is the way why the expression on the left is used in so many contexts.
Consider a sequence of function $$ \delta_\epsilon(x)=\begin{cases}\frac{1}{2\epsilon},&-\epsilon<x<\epsilon,\\ 0,&\mbox{otherwise}. \end{cases} $$
It is reasonable to expect that $$ \delta(x)=\lim_{\epsilon\to 0}\delta_\epsilon(x). $$
Now consider the integral and assume that we are free to change the order of operations: $$ \int_{\mathbb R}\delta(x)f(x)dx=\lim_{\epsilon\to 0}\int_{\mathbb R}\delta_{\epsilon}(x)f(x)dx=\lim_{\epsilon\to 0}\int_{-\epsilon}^{\epsilon}\frac{1}{2\epsilon} f(x)dx=\lim_{\epsilon\to0}2\epsilon\cdot \frac{1}{2\epsilon} f(\xi), $$ where due to the mean value theorem $\xi\in(-\epsilon,\epsilon)$.
Hence we can conclude that $$ \lim_{\epsilon\to 0}f(\xi)=f(0), $$ which gives you a "proof" of the original formula.