I learned two things in elementary statistics class - the concept of continuous random variable and probability of a point in continuous random variable is $0$ because the area is $0$ over the point (geometric understanding of probability?).
But when we are talking about (standard) normal distribution, it has a functional form of pdf $f(x)=\frac{1}{\sqrt{2 \pi}} e^{-\frac{1}{2}\left(x\right)^2}$ and this value can be "evaluated" at any point, for example, if $x=0$ then $f(0) = \frac{1}{\sqrt{2 \pi}}$ which is not $0$.
So I am confused here, the point of normal distribution looks life definitely get assigned non-zero values of probability.
What kind of confusion do I have at this point?
To expand on Andrew Zhang's comment, the values of $f$ are probability densities not probabilities. In order to get probabilities you need to integrate over some interval, the probability that you select a value in $[a,b]$ is $$ \int_{a}^bf(x) \, dx, $$ and if $a=b$, then this is zero.
If you are interested in the probability of selecting a value close to $0$ (say in $[-\delta, \delta]$) the appropriate integral is $$ \int_{-\delta}^\delta f(x) \, dx $$
Practically, you could use the PDF to estimate the probability that $X \in [x-\delta, x+\delta]$, we know it is approximately equal to $2\delta f(x)$ (for small enough $\delta$), i.e. that the probability of selecting a value in $[-\delta,\delta]$ is roughly $\frac{\sqrt{2} \delta}{\pi}$.