This seems very simple, that's why I want to check it.
Let X be an ordered set in the order topology. If X is connected, then X is a linear continuum.
The Attempt at a Solution
An ordered set is a set with an order relation "<" which is antireflexive, transitive, and satisfies, for any x, y, either x < y or x > y.
First of all, X cannot be finite. Since if it were, we could find two nonempty, disjoint open intervals whose union would equal X. So, X is infinite.
A linear continuum is a set with more than one element, with the least upper bound property, and the property that for any x, y in it, there is an element z such that x < z < y.
Clearly, an infinite set has more than one element.
Let x, y be two elements of X. If there were no element between x and y, X would be disconnected. So, there must be some element z such that x < z < y.
Let A be a non empty subset of X which is bounded above, i.e. for all x in A, x < M. If there were no elements of X between x and M, X would be disconnected. So, there exists some element x1 such that x < x1 < M. The same holds for x1 and M. Hence, for any neighbourhood of M (i.e. an open interval containing M), there exists some xn which is contained in the interval, which shows that M is the least upper bound for A.
Hence, X is a linear continuum.