quantum mechanics - Why is $langle x| x' rangle=delta(x-x')$?

I've tried to find any solution or proof for

$$\langle x| x' \rangle=\delta(x-x'),$$ but I only came to this post: Wave function and Dirac bra-ket notation

So I got the information, that the vector $|x\rangle$ form a dirac-normalized basis for the Hilbert Space.

I know that the dirac-delta distribution is defined like this:

$$\delta(x-x') = \begin{cases} 0 &\mbox{if } x\neq x' \\ \infty & \mbox{if } x=x' \end{cases},$$ this means that my x' is a point on my x axis where I have my infinite high peak. And also $$\int_{-\infty}^{\infty}dx\cdot \delta(x-x')=1.$$

But how actually correlate this with the scalar product of vectors x, x' in the Hilbert Space that form a so-called 'diracl-normailzed" basis of it?

Can you give me some tips on this please? Or maybe you actually know a link, where this is explained.

Answer

Isn't it just from the sifting property?

$$f(x) = \int\mathrm{d}x'\;f(x')\,\delta(x - x')$$

That is, if you accept the above and if you accept that

$$|\psi\rangle = \int \mathrm{d}x'\,\psi(x')\,|x'\rangle$$

then

$$\psi(x) = \langle x|\psi\rangle = \langle x| \int \mathrm{d}x'\,\psi(x') \,|x'\rangle = \int \mathrm{d}x'\,\psi(x')\,\langle x|x'\rangle$$

$$\Rightarrow \langle x|x'\rangle = \delta(x - x')$$