Why are these sets saturated open sets? 
How did the author find out that typical saturated open sets are those depicted? For me it's not clear why those are saturated open sets. How do I see it? For definition, see About saturated sets
Similarly, I can't see why these sets are saturated open:

 A: Remember that a saturated open subset $U \subset X$ with respect to a quotient map $p: X \rightarrow X^*$ is a set satistying $p^{-1}(p(U))=U$. So, if $x\in U$, every $y$ with $p(x)=p(y)$ must also be in $U$. 
In the first example,
1) If your saturated open $U$ contains an element of the boundary of the disk, it must contain the entire boundary, since they are all related, collapsed to one single point. Now, since it is open, it must be a union of open neighborhoods around each of its points. That's why it looks like that (it could also be the entire disk, not just a ring like in the picture).
2) If your saturated $U$ just contains elements of the interior of the disk, because the relation in the interior is trivial (i.e each element is related just to itself), it can be any open subset contained in the interior. Just with that it would be saturated. 
In the second example, your relation is to identify points in the boundary of the square if they are in the same vertical, or in the same horizontal line, and again in the interior don't identify anything.
1) If your saturated open subset just contains points in the interior, by the same reason as before it could be any open subset not touching the boundary.
2) If it intersects one of the sides and not any corner, for any segment of side it containd it would have to contain the corresponding segment in the opposite side. Of course, since it is open, it must be a neighborhood of both this segments, but its intersection with the interior could be any, there's no symmetry required here.
3) If it contains a vertex, it would have to contain all of them (the 4 vertices are identified) and of course neighborhoods of them such that any time it contains a segment of a side, it also contains the corresponding in the opposite side. But again its intersection with the interior could be any, no symmetry required.
