Reflexive Closure of Transitive Antisymmetric Relation is Ordering

Theorem
Let $S$ be a set.

Let $\prec$ be a transitive, antisymmetric relation on $S$.

Let $\preceq$ be the reflexive closure of $\prec$.

Then $\preceq$ is an ordering on $S$.

Antisymmetric
Let $a, b \in S$.

Suppose that $a \ne b$.

Then by the definition of the diagonal relation,
 * $\left({(a,b)}\right) \notin \Delta_S$ and
 * $\left({(b,a)}\right) \notin \Delta_S$.

Since $\prec$ is antisymmetric, $\left({(a,b)}\right)$ and $\left({(b,a)}\right)$ are not both in $\prec$.

Thus $a \not\preceq b$ or $b \not\preceq a$.