This is common sense. Suppose it is not the case that a given person is a student or is married. Is it the same as to say that this person is not a student or is not married?
Does the bar there mean topological closure? If so, then the answer is yes, they're equal. If the bar means set complement, then see Plato's response above.
I'll write the closure of a set X as as cl(X).
Suppose N is a closed set containing (A union B union C). Then N is a closed set containing A, so cl(A) contained in N. Likewise, cl(B) and cl(C) are contained in N. Thus (cl(A) union cl(B) union cl(C)) is contained in N. Since N was any closed set containing A, B, and C, have that (cl(A) union cl(B) union cl(C)) is contained in the intersection of all such closed sets. Thus (cl(A) union cl(B) union cl(C)) is a subset of cl(A union B union C).
Now notice that cl(a) is a closed set containing A, cl(B) is a closed set containing B, and cl(C) is a closed containing C. Thus (cl(A) union cl(B) union cl(C)) is a closed set that contains (A union B union C). Thus (cl(A) union cl(B) union cl(C)) contains cl(A union B union C).
Putting those two observations together proves that (cl(A) union cl(B) union cl(C)) = cl(A union B union C).
This obviously generalizes to any finite number of sets.