A scattered space if a space $X$ that contains no nonempty dense-in-itself subset. Equivalently, every nonempty subset $A$ of $X$ contains a point isolated in $A$.
Note that in general the union of two scattered sets is not scattered. For example, if $X=\{a,b\}$ with the indiscrete topology, $\{a\}$ and $\{b\}$ are both scattered, but their union, $X$, is not scattered as it has no isolated point.
It is shown in Kuratowski's book (p. 79) that in a $T_1$ space, the union of two scattered sets is scattered (he assumes spaces are $T_1$ unless mentioned otherwise). I think the following is more generally true:
Theorem: In a $T_0$ space, the union of two scattered sets is scattered.
How can this be proved?
Best Answer
This, and a number of similar results, are proven in Scattered spaces, compactifications and an application to image classification problem by M. Al-Hajri et. al.
The proof they give is roughly as follows: Let $A$ and $B$ be scattered subspaces, and $S\subseteq A\cup B$. We show that $S$ has an isolated point. Let $S_A \ = S\cap A$ and $S_B = S\cap B$, and note that $S_A$ and $S_B$ are scattered. Since $S_A$ is scattered, there is an $a\in S_A$ and and an open set $U$ such that $\{a\} = S_A\cap U$. If $U\cap S_B = \emptyset$, we're done. So assume not. Since $U\cap S_B \subseteq S_B$, it has an isolated point. So there exists $b$ in $U\cap S_B$ and an open set $V$ such that $\{v\} = U\cap S_B \cap V$.
Thus the open set $U\cap V$ is either $\{b\}$ (in which case we're done), or $U\cap V = \{a,b\}$, in which case we use the fact that our space is $T_0$ to isolate one of $a$ or $b$, completing the proof.