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)
Best Answer
If $V$, open subset of $\Bbb R^m$, were homeomorphic to an open subset of $\Bbb R^n$, $U$, let $f: U \to V$ be a homeomorphism. (Suppose WLOG that $m \leq n$.) Compose with a linear inclusion map $\Bbb R^m \hookrightarrow \Bbb R^n$ to get a continuous injective map $U \to \Bbb R^n$ with image contained in the subspace $\Bbb R^m$, as it factors as $U \to V \subset \Bbb R^m \hookrightarrow \Bbb R^n$.
If $m < n$, the image cannot be open - any neighborhood of a point in the hyperplane contains points not in the hyperplane. Therefore $m=n$. So invariance of domain implies invariance of dimension.
Of course, this says even more: there's not even a continuous injection $U \to V$ between open sets of $\Bbb R^n, \Bbb R^m$ respectively, $n>m$.