$A,B, C$ are independent $\implies (A \cup B)$ and $(B \cup C)$ are independent? Given that $A,B, C$ are independent events, I am trying to prove that $A \cup B$ and $B \cup C$ are independent.
$$P((A \cup B) \cap (B \cup C)) \\= P(B \cup (A \cap C)) \\= P(B) + P(A \cap C) - P(A \cap B \cap C) \\= P(B) + P(A)P(C) - P(A)P(B)P(C)\\ = P(B) + P(A)P(C) \left(1 - P(B) \right).$$
This clearly does not equal $P(A \cup B)P(B \cup C).$ How do I find an explicit counterexample to the claim that $A \cup B$ and $B \cup C$ are independent events? I am struggling because of the condition that $A,B,C$ must be independent.

Consider an experiment of tossing two coins where each outcome is uniform. We have $\Omega =$ {H,T}$^2$. Label events:
$A =$ Event that first coin toss is heads. $P(A) = \frac{1}{2}$
$B =$ Event that second coin toss is heads $P(B) = \frac{1}{2}$
$C = \emptyset$ $P(C) = 0$
Check that $A,B,C$ are independent.
$P(A \cap B) = P({HH}) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} $
$P(A \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $
$P(B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot 0 $
$P(A \cap B \cap C) = P(\emptyset) = 0 = \frac{1}{2} \cdot \frac{1}{2} \cdot 0 $
Now, check whether $A \cup B$ and $B \cup C$ are independent.
$P((A \cup B) \cap (B \cup C)) = P({HH,TH}) = \frac{1}{2}$. However, this is not equal to $P(A \cup B) \cdot P(B \cup C) = \frac{3}{4} \cdot \frac{1}{2} = \frac{3}{8}$, so not independent.
