Show that $\sigma(\mathscr{C}) = \sigma(\mathscr{G} \cup\mathscr{H})$ I'm doing some grad work as a past-time and I wanted to learn measure theory but I'm absolutely not managing it.

Let $\mathcal{G}$ and $\mathcal{H}$ be two $\sigma$-algebras on $\Omega$. Let $\mathcal{C} = \{G \cap H: G\in\mathcal{G}, H\in\mathcal{H}\}$.
Show that $\mathcal{C}$ is a $\pi$-system and that $\sigma(\mathcal{C})=\sigma(\mathcal{G \cup H})$.
Let $\mathcal{I} = \mathcal{G \cup H}$ and
$G,H \in \mathcal{I} \implies G\cap H \in \mathcal{I}$ 

then $\mathcal{I}$ is a $\pi$-system. 
From here on I'm lost.
 A: By definition, $\mathcal C$ is non-empty. Now: If $A,B\in\mathcal C$, then there are $G_1,G_2\in\mathcal G$, and $H_1,H_2\in\mathcal H$ such that $A=G_1\cap H_1$ and $B=G_2\cap H_2$, so $A\cap B=(G_1\cap G_2)\cap(H_1\cap H_2)\in C$, because $\mathcal G$ and $\mathcal H$ are $\pi$-systems. This proves that $\mathcal C$ is a $\pi$-system.
The key now is to note the obvious observation that if $\mathcal A\subset\mathcal B$, then $\sigma(\mathcal A)\subset\sigma(\mathcal B)$. We use this in what follows:
Note that $\Omega\in\mathcal G$ and $\Omega\in\mathcal H$, by definition of $\sigma$-algebra. It follows that if $A\in\mathcal G$, then $A=A\cap \Omega\in \mathcal C$, and similrly, if $B\in\mathcal H$, then $B=\Omega\cap B\in\mathcal C$. This means that $\mathcal C$ contains both $\mathcal G$ and $\mathcal H$, so it contains their union, and therefore $\sigma(\mathcal C)$ contains $\sigma(\mathcal G\cup \mathcal H)$.
On the other hand, $\sigma(\mathcal G\cup\mathcal H)$ is a $\pi$-system and contains as subsets both $\mathcal G$ and $\mathcal H$. In particular, it contains $\mathcal C$. But then it also contains $\sigma(\mathcal C)$.
A: $\mathcal{C}$ is $π$-system since $(G ∩ H) ∩ (G' ∩ H') = (G ∩ G') ∩ (H ∩ H') ∈ \mathcal{C}$. Obviously $\mathcal{C} ⊇ \mathcal{G} ∪ \mathcal{H}$ so $σ(\mathcal{C}) ⊇ σ(\mathcal{G} ∪ \mathcal{H})$. Also $\mathcal{C} = π(\mathcal{G} ∪ \mathcal{H}) ⊆ σ(\mathcal{G} ∪ \mathcal{H})$ so $σ(\mathcal{C}) ⊆ σ(\mathcal{G} ∪ \mathcal{H})$.
