I think your confusion is based on interpreting "either ... or" as meaning "one or the other but not both" (exclusive or). Although this interpretation is not uncommon in everyday English, that's not how I was using the phrase in the passage you quoted. I meant "or" in the usual mathematical-logic sense of "one or the other or both" (inclusive or).
So my claim on page 43 is that "every point of $M$ is either an interior point or a boundary point," but I'm not claiming (at this stage) that a point cannot be both. The fact that it cannot be both is what Theorem 2.59 (Invariance of the Boundary) guarantees, but the tools for proving it in full generality are not developed until Chapter 13 (Homology).
The issue of exclusive vs. inclusive or is a common source of confusion. In ordinary English, "either...or" is probably most often interpreted as exclusive or, but not universally. The meaning depends on context: in "you can have either soup or salad with your entree," it's clearly exclusive, but in "you must have either a bachelor's degree or three years' experience," it's just as clearly inclusive. The same problem arises even more with "or" alone (if you delete "either" from those two sentences, it doesn't change the meaning of either one).
While there may be mathematicians who use "either A or B" when they really mean "either A or B but not both," I think most careful mathematical writers insert some phrase such as "but not both" when they really mean exclusive or, and otherwise interpret "or" and "either...or" as inclusive. That's certainly the way I write. I'm sorry it confused you -- I'll try to be more sparing with my use of "either" from now on.
John M. (Jack) Lee
(the author)
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
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 $\phi(p) \in \partial \mathbb R^n_+ =\mathbb R^{n-1} \times \{ 0\}$.