Subset Relation is Ordering

Theorem
Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Let $\mathbb S \subseteq \mathcal P \left({S}\right)$ be any subset of $\mathcal P \left({S}\right)$, that is, an arbitrary set of subsets of $S$.

Then $\subseteq$ is an ordering on $\mathbb S$.

In other words, let $\left({\mathbb S, \subseteq}\right)$ be the relational structure defined on $\mathbb S$ by the relation $\subseteq$.

Then $\left({\mathbb S, \subseteq}\right)$ is a poset.

Proof
To establish that $\subseteq$ is an ordering, we need to show that it is reflexive, antisymmetric and transitive.

So, checking in turn each of the criteria for an ordering:

Reflexivity
So $\subseteq$ is reflexive.

Antisymmetry
So $\subseteq$ is antisymmetric.

Transitivity
That is, $\subseteq$ is transitive.

So we have shown that $\subseteq$ is an ordering on $\mathbb S$.