You are correct in pointing out that the definition of the Dirac delta is not that it equals zero at $\tau \neq 0$ and infinity at $\tau = 0$, but rather that it integrates to one over its entire range and effectively "samples" the function it is multiplied with at $\tau = 0$ when integrated, giving the notation \begin{equation} \int\delta(\tau)f(\tau)d\tau=f(0) \end{equation}
Another notation that I find helpful is to view the delta function from a centered normal distribution \begin{equation} \delta(x) \approx \lim_{\sigma\rightarrow 0} \frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-\frac{x^{2}}{2\sigma^{2}}} \end{equation} This approximation works since the integral of any PDF is \begin{equation} \int_{-\infty}^{\infty}f(x)dx = 1 \end{equation}
Practically speaking, in terms of proving that $x(t)$ is uncorrelated with $x(t+\tau)$, if you integrate $r_{x}(\tau)$ with any other function $f(\tau)$, you would get \begin{equation} \int r_{x}(\tau)f(\tau)d\tau = \int \frac{N_{0}}{2}\delta(\tau)f(\tau)d\tau = \frac{N_{0}}{2}f(0) \end{equation} thereby showing that $r_{x}(\tau)$ contributes a non-zero value only at $\tau = 0$, which shows that $x(t)$ is only correlated with a non-shifted version of itself.
If you want to delve further into the Dirac delta, you would have to delve into measure theory. However, further exploration into this topic is better asked on the Math SE, and I would not be qualified to give you that answer.