Let be $(X,\preccurlyeq)$ a totally ordered set. So if $Y$ is a bounded and closed subset of $X$ with respect the order topology then is it compact? if this is not generally true then is it true when $X$ has maximum of minimum?
Anyway it seem to that the statement is true when $X$ is a well ordered set with a minimum $m$ and maximum $M$ element, that is a well ordered space is compact.
Indeed if $\mathcal U$ is an open cover of $X$ then ther exist $U_0\in\mathcal U$ such that $$M\in U_0$$ However $U_0$ is a neighborhood of $M$ so that there must exist a basic neighrbohood of $M$ there contained, that is there must exist $x_0\in X$ such that $$(x_0,M]\subseteq U_0$$ Now if $\mathcal U$ cover $X$ then there exists $U_1\in\mathcal U$ such that $$x_0\in U_1$$ However $U_1$ is a neighrborhood of $x_0$ so that there must exist a basic neighrbohood of $x_0$ there contained, that is there must exist $x_1\in X$ such that $$(x_1,x_0]\subseteq U_1$$ So in this way we can define a sequence $(x_n)_{n\in\omega}$ in $X$ which has surely a minimum element and it seem to me that this element must be $m$ so that there exist $n\in\omega$ such that $x_n$ is $m$ and thus finally $U_0,\dots, U_n$ is a finite subcover of $\mathcal U$.
Anyway I point out that this do not prove that $Y$ is compact because the subspece topology of $X$ on $Y$ is not generally equal to the ordered topology on $Y$.
So could you explain why $m$ is the minimum of the above sequece? Moreover could you show is the statement is true also when $X$ is not well ordered, please?
Consider $X = [0,1) \cup (2,3]$ endowed with the usual order relation defined on $\mathbb{R}$. In the order topology, $[0,1)$ is a closed and bounded subset of $X$ that is not compact. In general, if $(X, \preceq)$ is a totally ordered set with the least upper bound property, then closed and bounded subsets of $X$ are compact in the order topology. But the space $X$ that I have defined here clearly does not have the least upper bound property, hence the counterexample. See Theorem 27.1 in Munkres for a proof of the statement about the least upper bound criterion.