This is obvious. Let $x \in M \setminus \partial M$. There exists an open neighborhood $V$ of $x$ in $M$ which is homeomorphic to $\mathbb R^n$. Then any $y \in V$ has the same property, hence $V \subset M \setminus \partial M$.
By the way, your definition of the boundary is not correct. Take $M = \mathbb R^n_+$. Then each $x \in \mathbb R^n_+$ has $\mathbb R^n_+$ as an open neighborhood which is homeomorphic $\mathbb R^n_+$. A correct definition is this: A point $p \in M$ is called a boundary point if there
exists a homeomorphism $\phi : U \to \mathbb R^n_+$ defined on an open neighborhood $U$ of $p$ in $M$ such that $\phi(p) \in \partial \mathbb R^n_+ =\mathbb R^{n-1} \times \{ 0\}$.
As mentioned in answers and comments, the definition given for a boundary point is not the correct one. A possible one should read:
A point $p \in M$ is a boundary point if it is such that $\pi_n(\phi(p))=\phi_n(p)=0$ for some chart $\phi:U \to \mathbb{R}^n_{+}$.
Then, I think, you would be willing to prove that if this is true for one chart $\phi: U \to \mathbb{R}^n_+$ as per the definition above, then it is true for any such chart. As one other answer suggests and references, this is not a big deal in the smooth setting.
This is true in the topological setting too, but due to a more involved reason: namely, the invariance of domain theorem. I don't know a reference which manages to avoid this.
Now, moving forward: Indeed, let $\psi: V \to \mathbb{R}^n_+$ be another chart. What we must prove then is that $\psi_n(p)=0$.
Instead of doing that, let's prove that if $\phi_n(p)>0$, then $\psi_n(p)>0$. It is easy to see (exchange places of $\phi$ and $\psi$) that this will prove that $\phi_n(p)>0$ if and only if $\psi_n(p)>0$, thus concluding that $\phi_n(p)=0$ if and only if $\psi_n(p)=0$.
Thus, suppose $\phi_n(p)>0$. We have that $\psi \circ \phi^{-1}$ is a continuous injection from $U \cap\mathbb{R}^n_+$ to $\mathbb{R}^n_+$, where $U$ is an open subset of $\mathbb{R}^n$. Restricting, $\psi \circ \phi^{-1}|_{U \cap \mathbb{R}^n_{>0}}: U \cap \mathbb{R}^n_{>0} \to \mathbb{R}^n$ is then injective and continuous. By invariance of domain, we know that such an image is open in $\mathbb{R}^n$. But this image is inside $\mathbb{R}^n_+$, therefore it can't intersect the set $\{x \mid x_n=0\}$. Since $\psi(\phi^{-1}(\phi(p)))=\psi(p)$ is in such image, it follows that it can't be such that $\psi_n(p)=0$.
As a sidenote which may also be of interest, a similar use of invariance of domain also yields that if $p$ is a boundary point, then there is no chart $\phi: U \to \mathbb{R}^n$ around $p$ (which is a homeomorphism with $\mathbb{R}^n$).
Best Answer
For 1, yes you have an $E$ around $\infty$ with $E$ homeomorphic to a ball, but you don't know anything about $\overline{E}$. In fact, $\overline{E}$ is potentially all of $M^\ast$!
On the other hand, inside of $E$ is another open set $E'$, also homeomorphic to a ball, but whose closure is homeomorphic to $\overline{\mathbb{B}^n}$. Use $U = M^\ast \setminus \overline{E'}$.
For 2, since $M\setminus U$ is homeomorphic to $\mathbb{R}^n\setminus \mathbb{B}^n$, it follows that $M\setminus\overline{U}$ is homeomorphic to $\mathbb{R}^n\setminus \overline{\mathbb{B}^n}$. Call such a homeomorphism $g$.
By using the inversion map $f$, one sees that $\mathbb{R}^n\setminus\overline{\mathbb{B}^n}$ is homeomorphic to $\mathbb{B}^n \setminus\{\vec{0}\}$.
Composing $g$ and $f$, we have a homeomorphism between $M\setminus\overline{U}$ and $\mathbb{B}^n\setminus \{\vec{0}\}$. Try to prove that we can use these to find a homeomorphism between $M^\ast \setminus \overline{U}$ and $\mathbb{B}^n$.