 Solution

The statement is false. Take two independent fair bits $U,V$ on the four equally probable outcomes $00,01,10,11$, and put $A_1=\{U=0\}$, $A_2=\{V=0\}$, and $B=\{U=V\}$. Each of $A_1,A_2,B$ has probability $1/2$, andThus $B$ is independent of each event separately. But $A_1\cup A_2$ contains $00,01,10$, whereas its intersection with $B$ contains only $00$. ThereforeThis is a concrete failure of extending separate independence to the generated union. A counterexample at the permitted value $n=2$ disproves the universal claim; for any larger $n$ one can repeat $A_1$ as the additional events.