Q: If $K$ is compact, prove that $K$ has a minimum and maximum. Let $X$ be a nonempty totally ordered set which doesn't have a minimum or maximum and let $T$ be order topology on $X$. Let $ K \subset X $ be a nonempty subset of $X$. If $K$ is compact, prove that $K$ has a minimum and maximum.
I tried by contradiciton. Suppose that $K$ has no maximum. Then $  <- \infty, k> k \in K$ is an open cover of $K$ which has no finite subcover. So K cannit be compact. Is this correct? 

A: Yes, your proof is correct, assuming that $\langle -\infty,k\rangle=\{x\in K:x<k\}$.
Indeed, if $x\in K$, $x$ is not the maximum, so there exists $k\in K$ with $x<k$; thus $x\in\langle -\infty,k\rangle$ and therefore
$$
\{\langle -\infty,k\rangle:k\in K\}
$$ is an open cover of $K$. It has no finite subcover, because
$$
\langle -\infty,k_1\rangle\cup\langle -\infty,k_2\rangle\cup\dots\cup
\langle -\infty,k_n\rangle=\langle -\infty,\max\{k_1,\dots,k_n\}\rangle
$$
which cannot be the whole of $K$, as it doesn't contain the upper bound.
