Upper Closure is Upper Set
Let $(S, \preceq, \tau)$ be an ordered set.
Let $T$ be a subset of $S$.
Let $U$ be the upper closure of $T$.
Then $U$ is an upper set.
Let $a \in U$.
Let $b \in S$ with $a \preceq b$.
By the definition of upper closure, there is a $t \in T$ such that $t \preceq a$.
By transitivity, $t \preceq b$.
Thus, agin by the definition of upper closure, $b \in U$.
Since this holds for all such $a$ and $b$, $U$ is an upper set.