I follow up on the previous post to report that the question stated there was answered by Dejan Govc. I’ll try to put together a self-contained version of his proof.

Let be a connected topological space. Fix a base point and define the *separation order* on as follows: iff the points and lies in different components of .

**Theorem**: *at most one of and holds. *

**Proof**. Suppose . Let be the connected component of in . Note that . If we can show that is connected, then and are in the same component of , which is to be proved. Suppose to the contrary that has a nonempty proper closed-open subset . We may assume . Note two things about :

- , because is closed in
- is open in . Indeed, open in . The closure of is contained in and therefore does not meet . Hence is open in , and therefore in .

Since , the set is **not** connected. So it has a nonempty proper closed-open subset . Then either or ; by replacing with its complement in we may assume the latter holds: that is, . We claim that is both closed and open in , contradicting the connectedness of . Indeed,

- is closed in and . Hence is closed in , and in .
- is open in and , hence is open in . Since is open in , it follows that is open in .

Nice exposition! The notation gives is very suggestive. I think there's a slight problem, though: I'm not sure is necessarily closed in .

I think it is quite easy to get rid of this problem though: is closed in which is closed in ; and since , we have that is open in . I hope this helps. Best regards, Dejan