Dedekind-Complete Bounded Ordered Set is Complete Lattice

Theorem
Let $\left({L, \preceq}\right)$ be an ordered set.

Suppose that $L$ has a lower bound $\bot$ and an upper bound $\top$.

Suppose also that $\left({L, \preceq}\right)$ is Dedekind-complete.

Then $\left({L, \preceq}\right)$ is a complete lattice.

Proof
Let $S \subseteq L$.

If $S = \varnothing$, then $S$ has a supremum of $\bot$ and an infimum of $\top$.

Suppose that $S \ne \varnothing$.

$S$ is bounded above by $\top$, so since $\left({L, \preceq}\right)$ is Dedekind complete, $S$ has a supremum.

$S$ is bounded below by $\bot$, so by Dedekind Completeness is Self-Dual, $S$ has an infimum.

Thus we have shown that every subset of $L$ has a supremum and an infimum, so $\left({L, \preceq}\right)$ is a complete lattice.