Here is a counterexample. Since this question is more restrictive than the sequel, it is a counterexample to that, too.

Let $\{1,2,3,4\}$ have equal probability.

Let $1 \in A_{1,n},A_{2,n}$.

Let $2 \in A_{1,2k},A_{2,2k+1}$.

Let $3 \in A_{1,2k+1},A_{2,2k}$.

Then $A_{1,n}$ is independent of $A_{2,n}.$ Each has probability $1/2$ and the intersection has probability $1/4$.

In the tail $\sigma$-algebra, $CI_1 =CI_2 = \{1\}$, and $CS_1=CS_2 = \{1,2,3\}$, and these are not independent.

Here is a counterexample. Since this question is more restrictive than the sequel, it is a counterexample to that, too.

Let $\{1,2,3,4\}$ have equal probability.

Let $1 \in A_{1,n},A_{2,n}$.

Let $2 \in A_{1,2k},A_{2,2k+1}$.

Let $3 \in A_{1,2k+1},A_{2,2k}$.

Then $A_{1,n}$ is independent of $A_{2,n}.$ Each has probability $1/2$ and the intersection has probability $1/4$.

In the tail $\sigma$-algebra, $CI_1 =CI_2 = \{1\}$, and $CS_1=CS_2 = \{1,2,3\}$, and these are not independent.