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.