Proofs of dirac delta property

Let $\mu = x-b$. Then you have that $$ \begin{align*} \int_{-\infty}^{\infty}\delta(a-x)\delta(x-b)dx &= \int_{-\infty-b}^{\infty-b}\delta(a-b-\mu)\delta(\mu)d\mu \\ & = \int_{-\infty}^{\infty}\delta(a-b-\mu)\delta(\mu)d\mu \\ & = \delta(b-a) \end{align*} $$

The '=' comes from the contraction of Dirac Functions. This result show that the contraction of two continuous Dirac functions is equivalent to a third Dirac function.

To be completely rigorous, note that in general the product of two measures/distributions/generalised functions is not well-defined; but the convolution is, which is what we have in this case (the convolution of $\delta(x)$ with $\delta(x-b)$ "evaluated" at $a$). Since the result is a measure, the equality just means that the integral against any function is the same, so we check this:

$$\int f(a) \int \delta(a-x) \delta(x-b)\ dx\ da = \int\left(\int f(a)\delta(a-x) da\right) \delta(x-b) dx \\ =\int f(x) \delta(x-b) dx=f(b)=\int f(a)\delta(b-a) da.$$

The interchange of integration here is essentially the definition of the convolution of measures - $\int \delta(a-x) \delta(x-b) dx$ is defined to be the measure such that this manipulation is valid.

I presume you know that $$\int f(x)\delta(x-a)\; dx = f(a).\tag{1}$$

Then put $f(x) = \delta(x-b)$ and you get $$\int \delta(x-b)\delta(x-a)\; dx = \delta(a-b).$$

Maybe this doesn't work because (1) applies only for well-behaved $f$? I forget. But if not you could make some limit argument based on $\delta(x) = \lim_{a\to\infty} \frac1{\sqrt a}e^{-x^2/a^2}$ and it would go through all right.