Show that $\mathcal{G} := \bigcap_{\theta\in\Theta}\mathcal{F}_{\theta}$ is also a $\sigma$-algebra.

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Let $\{\mathcal{F}_{\theta}:\theta\in\Theta\}$ be a family of $\sigma$-algebras on $\Omega$. Show that $\mathcal{G} := \bigcap_{\theta\in\Theta}\mathcal{F}_{\theta}$ is also a $\sigma$-algebra.

MY ATTEMPT

Since $\Omega\in\mathcal{F}_{\theta}$ for each $\theta\in\Theta$, we conclude that $\Omega\in\mathcal{G}$. Moreover, let $A\in\mathcal{G}$. Then $A\in\mathcal{F}_{\theta}$ for every $\theta\in\Theta$. Consequently, $A^{c}\in\mathcal{F}_{\theta}$ for every $\theta\in\Theta$. That is to say, $A^{c}\in\mathcal{G}$. Finally, one has that \begin{align*} A_{n}\in\mathcal{G},\,n\geq 1 & \Rightarrow A_{n}\in\mathcal{F}_{\theta},\,n\geq 1,\,\theta\in\Theta\\\\ & \Rightarrow \bigcup_{n=1}^{\infty}A_{n}\in\mathcal{F}_{\theta},\,\theta\in\Theta\\\\ & \Rightarrow \bigcup_{n=1}^{\infty}A_{n}\in\mathcal{G} \end{align*}

and we are done.

Is the wording of my proof correct?

Is there another way to prove? I am new to this so any contribution is appreciated.