Let $X$ be an ordered set in the order topology. Show that if $X$ is connected, then $X$ is a linear continuum.
My attempt:
Approach(1): first we show second property of linear continuum. Assume towards contradiction, i.e. $\exists x,y\in X$ with $x\lt y$ such that $(x,y)=\{ z\in X| x\lt z \lt y\}=\emptyset$. Then $(x,+\infty)$ and $(-\infty ,y)$ form separation of $X$. $(x,+\infty), (-\infty ,y)\in \mathcal{T}_X$. $(x,+\infty)\cap (-\infty ,y)=(x,y)=\emptyset$. $(x,+\infty)\cup (-\infty ,y)=X$. $y\in (x,+\infty)$ and $x\in (-\infty ,y)$. Which contradicts our initial assumption of $X$ being connected.
We can also use $[y,+\infty)$ and $(-\infty ,x]$ as separation of $X$. Because $[y,+\infty)$ and $(-\infty ,x]$ are closed in $X$. Using $[y,+\infty) \cap (-\infty ,x]=\emptyset$, $[y,+\infty) \cup (-\infty ,x]= X$ and $[y,+\infty), (-\infty ,x]\neq \emptyset$ facts, $[y,\infty)$ and $(-\infty ,x]$ are open in $X$.
Now we show $X$ have least upper bound property. Assume towards contradiction, $\exists A\subseteq X$ such that $A\neq \emptyset$ and bounded but $\sup(A)\notin X$. Since $A$ is bounded, $\exists z\in X$ such that $a\leq z, \forall a\in A$. Note equality don’t hold because $\sup(A)\notin X$. So $a\lt z, \forall a\in A$. By definition of $\sup$, $\sup (A)\lt z$. Since $A\neq \emptyset$, $\exists a\in A$ such that $a\lt \sup(A) \lt z$. Thus $(-\infty, \sup(A))$ and $(\sup(A),+\infty)$ form separation of $X$. $(-\infty, \sup(A)) ,(\sup(A),+\infty)\in \mathcal{T}_X$. $a\in (-\infty, \sup(A))\neq \emptyset$ and $z\in (\sup(A),+\infty)\neq \emptyset$. $(-\infty, \sup(A)) \cap (\sup(A),+\infty)=\emptyset$. $(-\infty, \sup(A)) \cup (\sup(A),+\infty)=X$. Thus we reach contradiction. Is this proof correct?
I’m not sure about second part of the proof because $\sup(A)\notin X$. So we are comparing points of $X$ with $\sup(A)\notin X$, i.e. $x\lt \sup(A)$, where $x\in X$.
Approach(2): https://math.stackexchange.com/a/1268096/861687.