Compact ordered topological space As a question inspired by this thread. I was just wondering whether a compact ordered topological space, a space with the total order topology, has to have a maximum and a minimum?
I was thinking that if we consider a collection consisiting of open rays of the form $\mathcal{C}=\{ x\in X:x<y \}$.
Assume $X$ is unbounded from above, then $\mathcal{C}$ would be a cover since we have a total order. 
By compactness of $X$ and since the rays are open, there exists a finite subcover of the form:
$$ \mathcal{C}'=\Big\{ \{x\in X:x<y_k\}  \Big\}_{k=1}^n $$
Because $<$ is a total order, we can assume WLOG that $y_1<y_2<...<y_n$, and in particular:
$$ \cup_{k=1}^n \{x\in X:x<y_k\} =\{x\in X:x<y_n\}  $$
And therefore $\{x\in X:x<y_n\}=X $. But $y_n\not<y_n$, so this is a contradiction. So we have a maximal element. 
In a similar argument with rays in the other direction, we have a minimal element. 
Do these steps seem valid, and can we then use this to generalize the maximum principle in the aforementioned post?
 A: Let S be a compact linear ordered topological space (LOT).
If S is not empty, then S has a maximum element.
Proof,  Assume otherwise.
Thus for all x in S exists y in S with x < y.
D(a) = { x in S : x < a } is open.
Show { D(x) : x in S } covers S.
As S is compact, there is a finite subcover,
Show that is contradictory.  
Likewise the order dual theorem compact LOTS have a minimum element.  
A: 
A non-empty LOTS (linearly ordered topological space) $(X,<)$ is compact iff every subset $A \subseteq X$ has a supremum $\sup(A)$. 

I show that in this post, so I won't repeat that proof here; 
This includes $\sup(\emptyset)=\min(X)$ and $\sup(X)=\max(X)$. 
But the minor corollary has an easy enough proof, as you showed:
Let $X$ be compact, and suppose it has no maximum. Let $L(x)=\{y \in X: y < x\}$ for all $x \in X$, which are open sets by the definition of the order topology.
When $X$ has no maximum, these sets form an open cover of $X$ (if $x \in X$, there must be some $x' > x$, and then $x$ is covered by $L(x')$), so finitely many, say $\{L(x): x \in F\}$ are a finite subcover, for some finite $F \subseteq X$. But then $m=\max(F)$ (exists in any linearly ordered set as $F$ is finite) cannot be covered by them, contradiction: there is no finite subcover and this contradicts compactness. So $\max(X)$ must exist.
The argument for $\min(X)$ is almost the same, but using the other subbasic elememts $U(x)=\{y \in X: x > y\}$ as the open cover and $\min(F)$ for the contradiction.
