Unit Disk Regular Surface?

To case 1),according to the definition, if $p\in\partial B$,where B is the closed disk,then for any neighborhood $V$ of $p$,$V\bigcap B$ is not an open set in $R^3$,so you cannot find any continues function $x:U\rightarrow V\bigcap B$ according to the topological definition of a continues function,where $U$ is any open set in $R^2$

To case 2),you can choose the map as $id_{R^2}$

picking any connected neighborhood of point $(0,1,0) \in \mathbb R^3$, we can show that the intersection with the closed disk $\overline D = \{(x, y, z) \in \mathbb R^3 | z = 0, x^2 + y^2 \le 1\}$ is not homeomorphic to an open set in $\mathbb R^2$.

the set of intersection is simply connected even if the point $(0, 1, 0)$ is removed, while any open set in $\mathbb R^2$ cannot be simply connected after removing a point from it.

notice that the image of embedding from $\mathbb R^2$ into $\mathbb R^3$ is always not open. in fact the property of being open can be changed depending on what space the set lies in.