Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact

Theorem
Let $X = \left[{{0}\,.\,.\,{1}}\right) \cup \left({{2}\,.\,.\,{3}}\right) \cup \left\{{4}\right\}$.

Let $\preceq$ be the ordering on $X$ induced by the usual ordering of the real numbers.

Let $\tau$ be the $\preceq$ order topology on $X$.

Let $Y = \left[{{0}\,.\,.\,{1}}\right) \cup \left\{{4}\right\}$.

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

Then:
 * $\left({Y, \preceq}\right)$ is a complete lattice.
 * $Y$ is closed in $X$.
 * $\left({Y, \tau'}\right)$ is not compact.

$Y$ is order-isomorphic to $\left[{{0}\,.\,.\,{1}}\right]$
Let $\phi: Y \to \left[{{0}\,.\,.\,{1}}\right]$ be defined by
 * $\displaystyle\phi\left({y}\right) =

\cases {y &\text{if }y \in \left[{{0}\,.\,.\,{1}}\right) \\       1 &\text{if }y = 4}$

Then $\phi$ is a order isomorphism.

Closed
If $x \in X \setminus Y$, then $x \in \left({{2}\,.\,.\,{3}}\right)$.

Thus $x \in \left({{\frac {x+2} 2}\,.\,.\,{\frac {x+3} 2}}\right)\in \tau'$.

Since the complement of $Y$ is open, $Y$ is closed.

Not Compact
Let $\mathcal A = \left\{{ {\downarrow}x: x \in \left[{{0}\,.\,.\,{1}}\right) }\right\} \cup \left\{{{\uparrow}2.5}\right\}$.

Then $\mathcal A$ is an open cover of $Y$ with no finite subcover.