Set Coarser than Upper Section is Subset

Theorem
Let $\left({S, \preceq}\right)$ be a preordered set.

Let $A, B$ be subsets of $S$ such that
 * $A$ is coarser than $B$

and
 * $B$ is an upper set.

Then $A \subseteq B$

Proof
Let $x \in A$.

By definition of coarser subset:
 * $\exists y \in B: y \preceq x$

Thus by definition of upper set:
 * $x \in B$