Compact Subspace of Linearly Ordered Space/Reverse Implication/Proof 1

Theorem
Let $\left({X, \preceq, \tau}\right)$ be a linearly ordered space.

Let $Y \subseteq X$ be a non-empty subset of $X$.

Suppose that the following hold:
 * $(1):\quad$ For every non-empty $S \subseteq Y$, $S$ has a supremum and an infimum in $X$.
 * $(2):\quad$ For every non-empty $S \subseteq Y$: $\sup S, \inf S \in Y$.

Then $Y$ is a compact subspace of $\left({X, \tau}\right)$.

Proof
Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.

Let $\preceq'$ be the restriction of $\preceq$ to $Y$.

Lemma
$(Y, \preceq', \tau')$ is a linearly ordered space.

Proof
By Equivalence of Definitions of Generalized Ordered Space/Two implies One and Equivalence of Definitions of Generalized Ordered Space/One implies Three:
 * $\tau'$ has a sub-basis consisting of upper and lower sets in $Y$.

To prove that $(Y, \preceq', \tau')$ is a linearly ordered space, we need to show that each element of this sub-basis is open in the $\preceq'$ order topology.

Let $U$ be a non-empty, $\tau'$-open upper set in $Y$.

If $U = \varnothing$ or $U = Y$, then $U$ is open in the $\preceq'$-order topology by the definition of topology.

Suppose then that $\varnothing \subsetneqq U \subsetneqq Y$.

Let $C = Y \setminus U$, so $C$ is nonempty and $\tau'$-closed.

$Y$ is $\tau$-closed by Closed Set in Linearly Ordered Space.

Thus $C$ is also $\tau$-closed.

$C$ has a supremum in $X$ by the premise.

Let $c = \sup_X C$

Since $C$ is $\tau$-closed, Closed Set in Linearly Ordered Space shows that $c \in C$.

Since $c \in C = Y \setminus U$ and $U$ is an upper set in $Y$:
 * $c \prec u$ for all $u \in U$.

Furthermore, if $y \in Y$ and $c \prec y$, then by the definition of supremum, $y \notin C$, so $y \in U$.

Thus $U = {\uparrow_Y}c$.

So $U$ is open in the $\preceq'$ order topology for $Y$.

The same argument shows that $\tau'$-open lower sets in $Y$ are open in the $\preceq'$ order topology.

Thus $\tau'$ is the $\preceq'$ order topology.

The premises immediately show that $(Y, \preceq')$ is a complete lattice.

By Complete Linearly Ordered Space is Compact, $Y$ is a compact subspace of $X$.