Power Set less Empty Set has no Smallest Element iff not Singleton

Theorem
Let $S$ be a set which is non-empty.

Let $\mathcal C = \mathcal P \left({S}\right) \setminus \varnothing$, that is, the power set of $S$ without the empty set.

Then the ordered structure $\left({\mathcal C, \subseteq}\right)$ has no smallest element $S$ is not a singleton.

Necessary Condition
Let $S$ not be a singleton.

Then $\exists x, y \in S: x \ne y$.

Let $Z \in \mathcal C$ be the smallest element of $\mathcal C$.

Then:
 * $\forall T \in \mathcal C: Z \subseteq T$

But by Singleton of Power Set less Empty Set is Minimal Subset, both $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are minimal elements of $\left({\mathcal C, \subseteq}\right)$.

Therefore it cannot be the case that $Z \subseteq \left\{{x}\right\}$ and $Z \subseteq \left\{{y}\right\}$.

Therefore $\left({\mathcal C, \subseteq}\right)$ has no smallest element.

Sufficient Condition
Let the ordered structure $\left({\mathcal C, \subseteq}\right)$ has no smallest element.

Aiming for a contradiction, suppose $S$ is a singleton.

Let $S = \left\{{x}\right\}$.

Then:
 * $\mathcal C = \left\{{\left\{{x}\right\}}\right\}$

Then:
 * $\forall y \in \mathcal C: y \subseteq \left\{{x}\right\}$

trivially.

Thus $\left({\mathcal C, \subseteq}\right)$ has a smallest element which is $\left\{{x}\right\}$.

By Proof by Contradiction it follows that $S$ is not a singleton.