Union of $\sigma$-algebras (Çınlar Exercise 1.18)
I agree with the first part of the proof. However, for the second, it is confusing to write $A = \cap_{j \in J} A_j$ and $B = \cap_{k \in K} A_{k}$ because (say $J=\{1,2\}$, $K=\{2,3\}$)), the set $A_2$ in the representation of $A$ and $B$ has no reason to be the same. Instead, we write $$ A' = \cap_{j \in J'} A'_j, \quad A'' = \cap_{k \in J''} A''_k $$ and we define $J=J'\cup J''$, $A_j=A'_j$ if $j\in J'\setminus J''$, $A_j=A''_j$ if $j\in J''\setminus J'$ and $A_j=A'_j\cap A''_j$ if $j\in J'\cap J''$. In this way, $$ A'\cap A''=\bigcap_{j \in J'\setminus J''} A'_j\cap \bigcap_{j \in J'\cap J''} A'_j\cap \bigcap_{k \in J''\setminus J'} A''_k\cap \bigcap_{k \in J'\cap J''} A''_k=\bigcap_{j\in J}A_j $$ each each $A_j$ belongs to $\mathcal E_j$.
I think Davide Giraudo has essentially solved the problem with my proof that $\mathscr C$ is a p-system, but looking at what I wrote, it's clear that my thinking was muddled, so below I've tried to fix it, taking inspiration from Davide Giraudo's answer.
If the sets $J$ and $K$ overlap at all---let's say $I = \{ 1, 2, 3, 4 \}, J = \{1, 2, 3 \}, K = \{ 2, 4 \}$---then the way I defined $A$ and $B$, we have $A = A_1 \cap A_2 \cap A_3, B = A_2 \cap A_4$. Notice that they both have $A_2$ in their respective intersections. That doesn't have to be true---they don't need to take the same set from $\mathscr E_2$. Perhaps I meant to define $B = \cap_{k \in K} B_{k}$. So let's try again.
Consider finite subsets $J, K \subseteq I$ and sets $A = \cap_{j \in J} A_j$ and $B = \cap_{k \in K} B_{k}$ where $A_j \in \mathscr E_j$ for $j \in J$, and $B_{k} \in \mathscr E_{k}$ for $k \in K$. Consider the intersection $A \cap B$. We need to write this in a way that complies with the rules of $\mathscr C$. This intersection will be over $J \cup K$ because we're combining the two intersections. We want to write $A \cap B = \cap_{l \in J \cup K} C_l$ but we need to figure out how to define the sets $C_l$ so that $A \cap B \in \mathscr C$. There are three possibilities for the sets $C_l$.
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If $l \in J, l \in K$ then define $C_l := A_l \cap B_l$.
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If $l \in J, l \notin K$ then define $C_l := A_l$.
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If $l \notin J, l \in K$ then define $C_l := B_l$.
So according to this definition, $C := A \cap B = \cap_{l \in J \cup K} C_l$ is an intersection over a finite subset $J \cup K \subseteq I$, and for each $l \in J \cup K$, we have $C_l \in \mathscr E_l$. This is clear in cases 2 and 3 above, and in case 1, since $A_l, B_l \in \mathscr E_l$, by the rules of a $\sigma$-algebra, we know that $A_l \cap B_l \in \mathscr E_l$. Thus $C = A \cap B \in \mathscr C$ and we conclude that $\mathscr C$ is a p-system.