Recall first the definition of a regular surface:
A subset $S\subseteq\mathbb{R}^3$ is called a regular surface if for every point $p\in S$, there is an open ball $B_r(p)$ around $p$ and an open set $A\subseteq\mathbb{R}^2$ and a continuously differentiable function $f:A\to S\cap B_r(p)$ such that $rank(Df) = 2$ and $f$ is surjective.
To my understanding, the notion that captures the fact that surfaces are '2 dimensional' in some intuitive sense, is the fact that $rank(Df) = 2$, and that $A\subseteq \mathbb{R}^2$, so in some intuitive sense, the image of $f$ cannot be '3 dimensional'.
Given that, it should be quite intuitive that regular surfaces must be of measure zero ('2 dimensional' object in 3 dimensional world). I have had many failed attempts, including attempts to construct some Lipschitz function and to find some set of measure zero that will map to the surface (since Lipschitz functions preserve zero measure). Although intuitive, it seems non-trivial.