Prove that two separated sets must be relatively open in their union I'm working in the plane here. My idea was that if A is open, then let O = A and then O intersect (A union B) = A, since A and B are separated, so A is relatively open in A union B. If A is not open, then there are some points in A that have neighborhoods not contained in A. I'll make an aggregate set called P of those points. Then, since A and B are separated, each of these points has a neighborhood containing no points in A. If I let O = (A union N_p) for all p in P, then O intersect (A union B) = A. Furthermore, O is open since each point in A\P is in O and has a neighborhood in A, and, thus, has a neighborhood in O, and each point in P has a neighborhood in O. Thus, A is still relatively open in A union B. 
My professor didn't think this was a valid solution. Can anyone help?
 A: Suppose that $A$ and $B$ are two separated sets in a space $X$ (no need to use anything specific to the plane). This means that $\overline{A} \cap B = \emptyset = A \cap \overline{B}$. In particular this means that $A$ and $B$ are disjoint. So to see that $A$ is relatively open in $C = A \cup B$, it is enough to see that its complement (in $C$!), which is $B$, is relatively closed in $C$, and the same holds (mutatis mutandis) for $B$.
So we need to see that $A$ and $B$ are relatively closed in $C$.
Now, in general, $\overline{A}^{(C)} = \overline{A} \cap C$, for every subset $C$ of $X$ and every $A \subset C$, and where the first term is the closure of $A$ in the relative (subspace) topology of $C$. Apply this to $C = A \cup B$: 
$$\overline{A}^{(A \cup B)} = \overline{A} \cap (A \cup B) = (\overline{A} \cap A) \cup (\overline{A} \cap B) = A,$$ because $A \subset \overline{A}$ and $A$ and $B$ are separated.    Hence, $A$ is closed in $C$ (and so $B = C \setminus A$ is open in $C$), and the same holds for $B$ again, interchanging the roles.
