Power Set $\sigma$-algebra

Proof

First note that $\varnothing \in \mathcal{P}(S)$ since $\varnothing \subseteq \mathcal{P}(S)$.

Let $A\subseteq S$. Then $A^c=S-A\subseteq S$ since by definition $S-A$ is $S$ with some subset removed.

Finally, consider a countable sequence of subsets of $S$:

Now let $x\in \bigcup _{i\in \mathbb{N}}{A}_i$. Hence by the definition of the union $x\in A_i$ for some $i\in \mathbb{N}$. However then by the definition of the subset, we have that $x\in S$.

Therefore

and hence