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Here is Prob. 4, Sec. 24, in the book Topology by James R. Munkres, 2nd edition:

Let $X$ be an ordered set in the order topology. Show that if $X$ is connected, then $X$ is a linear continuum.

And, here is the definition of linear continuum:

A simply ordered set $L$ having more than one element is called a linear continuum if the following hold:

(1) $L$ has the least upper bound property.

(2) If $x < y$, then there exists $z$ such that $x < z < y$.

My Attempt:

Since $X$ is a simply ordered set having the order topology, $X$ has more than one element, by the definition in Sec. 14 in Munkres.

(1) Let $A$ be a non-empty subset of $X$ such that $A$ is bounded above in $X$. Let $U$ be the set of all the upper bounds in $X$ of set $A$; that is, let $$ U \colon= \{ \ u \in X \ \colon \ a \leq u \mbox{ for every } a \in A \ \}.$$ This set $U$ is of course non-empty. If $U$ has a smallest element (or $A$ has a largest element, which would then be the smallest element of $U$), then that element would be the least upper bound of set $A$, and we are done.

So let us assume that $U$ has no smallest element. Then, for any element $u \in U$, there exists an element $u^\prime \in U$ such that $u^\prime < u$, and so $$ u \in \left( u^\prime, +\infty \right) \subset U,$$ with $\left( u^\prime, +\infty \right)$ being an open set in $X$. This shows that $U$ is open in $X$.

Now as $X$ is connected and as $U$ is a non-empty open subset of $X$, so $U$ cannot be closed. Therefore, $U$ has a limit point $b$, say, that does not belong to $U$. Then $b$ is not an upper bound of set $A$, which implies the existence of an element $a \in A$ such that $b < a$, we can also conclude that $$ b \in (-\infty, a) \subset X\setminus U,$$ with $(-\infty, a)$ being an open set. This contradicts our choice of $b$ as a limit point of set $U$. Hence the set $U$ of all the upper bounds in $X$ of set $A$ must have a smallest element, and that element is the least upper bound of $A$.

Thus we have shown that every non-empty subset $A$ of $X$ that is bounded above in $X$ has a least upper bound in $X$. Therefore, $X$ has the least upper bound property.

(2) Suppose that $a$ and $b$ are any two elements of $X$ such that $a < b$. If there is no element $c \in X$ such that $a < c < b$, then $X$ is the union of the open rays $$(-\infty, b) \colon= \{ \ x \in X \ \colon \ x < b \ \} $$ and $$ (a, +\infty) \colon= \{ \ x \in X \ \colon \ a < x \ \},$$ both of which are open sets in the order topology and are also non-empty; the former contains $a$, whereas the latter contains $b$. This contradicts the fact that $X$ is connected. So there exists an element $c \in X$ such that $a < c < b$.

Is this proof satisfactory enough?

Or else, is there any problem in the logic or presentation thereof? If so, where exactly is this proof in need of improvement or correction?

Finally, here is Theorem 24.1 in Munkres:

If $L$ is a linear continuum in the order topology, then $L$ is connected, and so are intervals and rays in $L$.

Can the (first part of) this theorem and the statement of Prob. 4, Sec. 24, be regarded as converses of each other?

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Very well done! There is only a small problem. That's when you write “$U$ is a non-empty open subset of $X$, so $U$ cannot be closed”. Yes, it can! But only in one case: that's when $U=X$. But you know that $U\varsubsetneq X$ (since $A\subset X\setminus U$ and $A\neq\emptyset$) and therefore that is not a problem.

And, yes, that problem and the theorem that you mention can be seen as converses of each other.

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