Subset of Set is Coarser than Set

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

Let $A, B$ be subset of $S$ such that
 * $A \subseteq B$

Then $A$ is coarser than $B$.

Proof
Let $x \in A$.

By definition of subset:
 * $x \in B$

By definition of reflexivity:
 * $x \preceq x$

Thus
 * $\exists y \in B: y \preceq x$