I have found these two posts that touch aspects of this question: "unions of $\sigma$-algebras" and this "unions of $\sigma$-algebras".
But I would like to provide a more comprehensive question. So, hopefully other users can benefit from it and any potential erros here can be corrected. This exercise is taken from exercise 1.18 in chapter 1 of "Probability and Stochastics" by Erhan Çinlar.
Exercise
Consider a set $E$ and let $\mathcal{E}_i$ be a $\sigma$-algebra on $E$ for all $i \in I$, where $I$ is an arbitrary index (possibly uncountable). Let $\mathcal{E}_I=\bigvee_{i \in I}\mathcal{E}_i=\sigma\big(\bigcup_{i \in I}\mathcal{E}_i\big)$. And define
\begin{equation} \mathcal{C}=\Bigl\{A \subset E: A=\bigcap_{i \in J} A_i \;\;\; \text{for some $J \subset I$, $J$ finite} \Bigr\} \end{equation} where $J \subset I$ is a finite subset of $I$.
Show that $\mathcal{E}_i \subset \mathcal{C} \; \forall i \in I$, so that $\bigcup_{i \in I}\mathcal{E}_i \subset \sigma \mathcal{C}$, and show that $\sigma C= \mathcal{E}_I$. Also, show that $\mathcal{C}$ is a $\pi$-system.
My solution
I show the equality by showing both inclusions.
First, let us show that $\mathcal{E}_I \subset \sigma \mathcal{C}$. To this end, consider $A \in \mathcal{E}_{\bar{i}}$ for some $\bar{i} \in I$. Hence, $A=\bigcup_{i \in J}A_i$ with $J=\bigl\{ \bar{i} \bigr\} $, i.e., $A \in C$. This holds for any $A \in \mathcal{E}_i$ and for all $i \in I$. Hence $\mathcal{E}_i \subset \mathcal{C}$ for all $i \in I$. This implies that $\bigcup_{i \in I}\mathcal{E}_i \subset \mathcal{C}$, which in turn implies \begin{equation} \mathcal{E}_I = \bigvee_{i \in I} \mathcal{E}_i = \sigma \bigl( \bigcup_{i \in I} \mathcal{E}_i \bigr) \subset \sigma \mathcal{C} \end{equation}
Let us now show that $\sigma\mathcal{C} \subset \mathcal{E}_I$. Consider $A \in \mathcal{C}$. By definition, there is $J \subset I \;$, $J$ finite, such that $A=\bigcup_{i \in J}A_i$ , with $A_i \in \mathcal{E}_i \subset \mathcal{E}_I \; \forall i \in J$
Then \begin{equation} A_i \in \mathcal{E}_I \; \forall i \in J \implies \bigcup_{i \in J}A_i \in \mathcal{E}_I \end{equation} So, $A \in \mathcal{C} \implies A \in \mathcal{E}_I$ with $A$ arbitrary, hence $\mathcal{C} \subset \mathcal{E}_I$. So, $\mathcal{E}_I$ is a $\sigma$-algebra containing $\mathcal{C}$, which implies that $\mathcal{E}_I$ contains also the smallest $\sigma$-algebra containing $\mathcal{C}$, i.e., $\sigma \mathcal{C} \subset \mathcal{E}_I$.
Hence, $\sigma \mathcal{C} = \mathcal{E}_I$.
Lastly, the fact that $\mathcal{C}$ is a $\pi$-system is immediate. Consider $A_1, A_2 \in \mathcal{C}$. Then $A_1 = \bigcup_{i \in J_1}$ for some finite $J_1 \subset I$ and $A_2 = \bigcup_{i \in J_2}$ for some finite $J_2 \subset I$. Hence
\begin{equation} A_1 \cap A_2 = \biggl( \bigcap_{i \in J_1}A_i \biggr) \cap \biggl( \bigcap_{i \in J_2}A_i \biggr) = \bigcap_{i \in J_1 \cup J_2}A_i \in \mathcal{C} \end{equation}
because $J_1 \cup J_2$ is again finite.
So, $A,B \in \mathcal{C} \implies A \cap B \in \mathcal{C}$, i.e., $\mathcal{C}$ is a $\pi$-system.
I think my solution is correct. If anyone sposts any errors, I would be more than happy to read them!