This community wiki solution is intended to clear the question from the unanswered queue.
The answer was given in the comments by Moishe Kohan and Lee Mosher.
Certainly Alexander Duality is needed in a form covering arbitrary compact subsets of $\mathbb R^n$. This involves Čech cohomology $\check{H}^*$ and states that for any compact $K \subset \mathbb R^n$ there exists an isomorphism
$$\tilde{H}_q(\mathbb R^n \setminus K) \approx \check{H}^{n-q-1}(K).$$
Here $\tilde{H}_*$ denotes reduced singular homology.
If $K, K'$ are homeomorphic, this implies $\tilde{H}_0(\mathbb R^n \setminus K) \approx \check{H}^{n-1}(K) \approx \check{H}^{n-1}(K') \approx \tilde{H}_0(\mathbb R^n \setminus K')$. In particular both complements must have the same number of path components. Since both complements are open in $\mathbb R^n$, they are locally path connected whence components and path components agree.
However, Alexander duality is much stronger than your generalization of the Jordan-Brouwer Theorem.
First, an important observation: Montiel and Ros seem to diverge from the more commonly accepted terminology of "closed manifold" to mean "compact and without boundary". By closed they really seem to mean closed in the topological sense, and manifold/surface is already assumed to be boundaryless.
They prove the Jordan-Brouwer separation theorem (for closed surfaces in $\mathbb{R}^3$) assuming compactness, and leave the closed case as an exercise. This is the exercise in the question.
With that said, for $(b)$, you can perturb the triangle generated by $p_1,p_2,p_3$ a little in order for it to be transverse to the surface and not change the number of intersections of the lines and the surface. Now, the intersection of such a triangle with the surface is a compact $1$-manifold with boundary and its boundary is precisely the intersections of the lines $[p_1,p_2], [p_2,p_3]$ and $[p_1,p_3]$ with the surface. Of course, $[p_2,p_3]$ does not intersect the surface, so it counts only the intersections of the lines $[p_1,p_2]$ and $[p_1,p_3]$. Since a compact $1$-manifold with boundary is a union of circles and closed intervals, the boundary has an even number of points and the result follows.
For $(c)$, you can pretty much follow the argument he gives for the compact case and define the two sets $A_{even}$ (resp. $A_{odd}$) as the set of points $x$ of $\mathbb{R}^3-S$ such that there exists a segment starting at $x$, ending in $B$ and intersecting $S$ an even (resp. odd) amount of times. Both sets are open by basic facts of transversality, and both sets are disjoint by $(b)$. $A_{even}$ is non-empty since you can pick a point in the ball and have empty intersection. That $A_{odd}$ is non-empty follows from the fact that if you consider a minimum of the function $x \mapsto \lVert x-p_0\rVert$ on the surface then you just need to extend the segment connecting $p_0$ to the minimum a bit in order to get an element of $A_{odd}$. By the way, such a minimum exists because the minimum of the function restricted to the intersection of the surface with an appropriately large enough closed ball centered on $p_0$, thus a compact set where the existence of the point achieving a minimum is guaranteed, is the same as the minimum of the function on the entire surface.
Now you know that $\mathbb{R}^3-S$ is not connected. Then Proposition $4.4$ in the book concludes what is requested at the end. For reference, the proposition states:
Let $S$ be a closed connected surface. Then $\mathbb{R}^3-S$ has at most two connected components. Moreover, each connected component of $\mathbb{R}^3-S$ has $S$ as its boundary.
Since the arguments therein are not relevant to the exercise and take up a page, I don't see it being much useful to transfer them here. But this ends the exercise.
Best Answer
Yes, this argument works. Jordan-Brouwer states that any homeomorph of $S^n$ in $\Bbb R^{n+1}$ separates it. There is, as you say, a homeomorph $S^m$ ($m<n$) in $\Bbb R^{n+1}$ which does not separate it. So $S^m$ is not homeomorphic to $S^n$.