Compact Spaces are Complete
# Statement 1
Let $(K, \leq)$ be a Total Ordering equipped with the Order Topology that is Compact. Then it is a Complete Ordering.
# Proof 1
Let $A \subset K$ be Nonempty and bounded from above. Consider $A$ as a monotone Net, $(a_{a}){a \in A} = A$. Because $K$ is Compact and A Set is Compact iff all Nets have a Convergent Subnet, we know there exists a Cofinal Subset $S \subset A$ so that $(b{b}){b \in S}$ is a convergent Subnet. It is montone because its parent is. Because Order-Preserving Nets Converge to their Supremum, we know $b{b} \to \sup\limits_{b \in S} b_{b} = \sup\limits S$. Because Supremum of a Cofinal Subset is the Supremum of the Set, it follows that $\sup\limits S = \sup\limits A$ and $\sup\limits A$ exists. $\blacksquare$
# Proof 2
We prove by Contraposition. Suppose there exists an $A \subset K$ Nonempty and bounded from above so that $\sup\limits A$ does not exist. Then, letting $U = {b \in X : b > a \text{ } \forall a \in A}$, $${(\leftarrow, a) : a \in A} \cup U$$ forms a Set Cover of $K$. This follows because for $x \in K$, either there exists $a \in A$ so that $a > x$ or there does not. In the former case, $x \in (\leftarrow, a)$. In the latter case, $x$ is an Upper Bound of $A$, and since $\sup\limits A$ does not exist, we cannot have $x = a$ for any $a \in A$. Thus $x \in U$. $U$ is also Open since for any $b \in U$, there exists $c \in U$ so that $c < b$ and $(c, \rightarrow) \subset U$. The $c \in U$ exists because $b$ is an Upper Bound that is not a Supremum. Thus our Set Cover is an Open Cover.
We claim that our Open Cover cannot be reduced to a finite Open Subcover. If it could, then we would have some $a_{1}, \dots, a_{n}$ (for some $n \in \mathbb{N}$) so that ${(\leftarrow, a_{i}) : i \in [n]} \cup U$ is an Open Cover of $K$. Since $A \cap U = \emptyset$, we must have $A \subset \bigcup\limits_{i = 1}^{n} (\leftarrow, a_{i}) = (\leftarrow, \max\limits_{i \in [n]} a_{i})$. But then $M = \max\limits(a_{1}, \dots, a_{n})$ is an Upper Bound of $A$ and $M \in A$ by construction. So $M = \sup\limits A$ $\unicode{x21af}$.
So we have that $K$ cannot be Compact. $\blacksquare$