Proof of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra

Claim 1. We check the three defining properties of Sigma-Algebra and Measurable Space for $\mathcal{I}$.

Every $\sigma$-algebra on $X$ contains $X$, so $X$ belongs to every member of $\mathcal{S}$ and hence $X\in\mathcal{I}$; when $\mathcal{S}$ is empty this holds vacuously, as do the two conditions below.

Let $A\in\mathcal{I}$ and let $\mathcal{G}\in\mathcal{S}$. Then $A\in\mathcal{G}$, so $X\setminus A\in\mathcal{G}$ because a $\sigma$-algebra is closed under complements. As $\mathcal{G}$ was an arbitrary member of $\mathcal{S}$, we get $X\setminus A\in\mathcal{I}$.

Let $(A_m)_{m\in\mathbb{N}}$ be a sequence in $\mathcal{I}$ and let $\mathcal{G}\in\mathcal{S}$. Then $A_m\in\mathcal{G}$ for every $m$, so $\bigcup_{m\in\mathbb{N}}A_m\in\mathcal{G}$ because a $\sigma$-algebra is closed under countable unions. Again $\mathcal{G}$ was arbitrary, so $\bigcup_{m\in\mathbb{N}}A_m\in\mathcal{I}$.

Claim 2. Let $\mathcal{S}$ be the collection of all $\sigma$-algebras on $X$ that contain $\mathcal{C}$. By Generated Sigma-Algebra the generated $\sigma$-algebra $\sigma(\mathcal{C})$ is the family of those subsets of $X$ that belong to every member of $\mathcal{S}$, so claim 1 applies to it and makes it a $\sigma$-algebra on $X$.

If $A\in\mathcal{C}$ then $A$ belongs to every member of $\mathcal{S}$, each of which contains $\mathcal{C}$; hence $A\in\sigma(\mathcal{C})$, and $\mathcal{C}\subseteq\sigma(\mathcal{C})$.

Finally let $\mathcal{G}$ be a $\sigma$-algebra on $X$ with $\mathcal{C}\subseteq\mathcal{G}$. Then $\mathcal{G}\in\mathcal{S}$, so every member of $\sigma(\mathcal{C})$ belongs to $\mathcal{G}$; that is, $\sigma(\mathcal{C})\subseteq\mathcal{G}$.