Is it true that any closed set of an arbitrary topological space X is the boundary of some subset of X?

In general this question has a negative answer: take a discrete topological space X and Z a non empty closed subset of X. Suppose Z=Fr(A) for some subset A of X, where Fr(A) denotes the boundary of A. Now since X is discrete, A is closed, which means Fr(A)⊂A but A is also open so A ∩Fr(A)=Ø which imply Fr(A)=Ø so that Z is empty but this is a contradiction.

Is the question affermative in ℝⁿ for n>0 ?

The answer is yes ! Indeed take Z a closed subset of ℝⁿ. Recall that ℝⁿ is a second numerable topological space (N2). Since Z is a subspace of a N2 topological space in particular Z is N2.But N2-space are N3-space (the proof can be seen for example in “Introduzione alla topologia generale” , Andrea Loi, pag 51). This means that there is S ⊂ Z dense and numerable. I claim that Fr(S)=Z. Indeed Int(S) is open, so union of open ball of ℝⁿ. A generic open ball of ℝⁿ is not numerable so that Int(S)=Ø. Now indicating with Closure(S) the closure of S and Est(S) the set of exterior point of S we get: Z=Closure(S) \ Est(S)=Z \ Est(S) which imply Est(S)=Ø. Then:

Z=Int(S)∪Fr(S)∪Est(S)=Fr(S).