Let $C$ be a figure "8" in th $xy$ plane and let $S$ be the cylindrical surface over $C$ that is,
$$S=\{(x,y,z)\in \mathbb{R}^3; (x,y)\in C\}.$$
Is the set $S$ a regular surface ?
So my answer is no because based on the proposition 2 in Do Carmo page 62 the image of local charts should be in this forms: $z=f(x,y)$ or $y=g(x,z)$ or $x=h(y,z)$, but the projection of $S$ in any of the planes doesn't have a one-to-one functions so $S$ in not a regular surface what do you think ?
It is indeed a regular surface. Recall that the figure 8 is a smooth manifold of dimension 1 and so is the real line. The product of two 1-(smooth)manifolds is a 2-(smooth)manifold, so in this case $S=C\times\mathbb{R}$, which gives you a smooth manifold of dimension 2, which is the same as a regular surface.
EDIT In the answer, I am implicity using a particular topology for the figure 8. If your definition of regular surface requires $S$ to be an euclidean subspace of $\mathbb{R}^3$, this reasoning is not valid, since the figure 8 doesn't admit a differentiable structure with euclidean topology. If this is the case, if you assume $S$ to be regular, then you could project it onto the 8 (the projection is a differentiable map), and by composition of differentiable maps, you would have a regular parametrization of the 8, which is a contradiction.