Jordan-Brouwer theorem for closed surfaces The following is Exercise (6) of Chapter 4 of Curves and Surfaces, 2nd edition, by Montiel and Ros, in which the reader is asked to prove the Jordan-Brouwer Theorem for closed surfaces

Theorem(Jordan-Brouwer): If $S$ is a closed connected surface, then $\Bbb{R}^3 \setminus S$ has exactly two connected components whose common boundary is $S$.

My question is in itens b and c below. How to prove them?
Thanks in advance and kind regards. 

EDIT Why isn't the following a counterexample to b?

 A: 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.
