**EDIT:** This is a geometric interpretation of the statement of the Krull Intersection Theorem, and not the proof, as the OP asked for. Here it is anyways.

Ok, so let's work in the setting of a Noetherian local ring $(R, \mathfrak{m})$. Then the Krull intersection says that $\bigcap_{n \geq 1} \mathfrak{m}^n = (0)$. (You can remove the local hypothesis if you assume that $R$ is a domain, and then $\mathfrak{m}$ can be any ideal).

The easiest geometric interpretation of this statement that I can think of is something like the following.

*There is no hypersurface passing through a point (or subvariety) of a variety $X$ with infinite order of vanishing through that point/subvariety.*

Or,

*The only function which vanishes to arbitrarily high order at a point is the zero function.*

The interpretation should be pretty easy. Given an $f \in R$, we can measure the order of vanishing of $f$ by asking what's the biggest power $n$ of $\mathfrak{m}$ such that $f \in \mathfrak{m}^n$ but such that $f \notin \mathfrak{m}^{n+1}$. Given now a scheme $X$, go look at a stalk of some (possibly non-closed) point.

Obviously, this result shows up a lot when studying completion, so it has geometric applications as well.