How about choosing $B = B_1 - B_2 $ and $C = C_1 -C_2$, we have $ B in S(E cap A)$ by property of $ S(E cap A) $, and $ C in S(E) $, by property of $S(E)$.
The fact that this works can be seen by observing that both $B1 cup (C1−A)$, and $B2cup (C2−A)$ are in fact both disjoint unions. $B_1,B_2$ has everything to do with $A$ and $C_1 -A, C_2 -A$ has nothing to do with $A$ at all.
ps: I don't know why you use the and and or symbol here instead of intersection and union symbol.
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