A set $S \subset \mathbb{R^3}$ is a regular surface iff, for every point $p$ in $S$ there is a function $\phi: U \subset \mathbb{R}^2 \to W = V \cap S$, called a parametrization of $p$, where $U$ is open in $\mathbb{R^2}$ and $V$ is open in $\mathbb{R^3}$, satisfying
1 - $\phi$ is $C^\infty$
2 - $\phi$ is a homeomorphism
3 - For every $X \in U$, the differential $dq_X: \mathbb{R^2} \to \mathbb{R^3}$ is injective.
I've read the following solution:
Consider $D = \{(x, y, z) \in \mathbb{R}^3 : |(x, y)| \leq 1 \land z = 0\}$. Assume, for contradiction, that $D$ is a regular surface. Let $p \in D$ with $|p| = 1$. Take a parametrization $\phi: U \subset \mathbb{R}^2 \to W \subset D$ of $p$. Without loss of generality, assume $\frac{\partial (x, y)}{\partial(u, v)} \neq 0$. Consider $\pi \circ \phi: U \to \mathbb{R}^2$, where $\pi: \mathbb{R}^3 \to \mathbb{R}^2$ is given by $(x, y, z) = (x, y)$.
By the inverse function theorem, there are neighborhoods $V_1$ and $V_2$ of $q = \phi^{-1}(p)$ and $\pi(p)$ respectively, such that $\psi = (\pi \circ \phi)|V_1: V_1 \to V_2$ is a homeomorphism. Thus, $D$ cannot be a regular surface because $V_2$ is not an open set in $\mathbb{R}^2$.
But it seems to me that there is no contradiction in $\psi(V_1) = V_2$ unless you use a stronger argument, such as the theorem of invariance of domain. So, I don't know how to prove that $D$ is not a regular surface, and the proof above seems wrong. Do you have any suggestions on how to prove it without appealing to algebraic topology?