A question about "rearranging" finite unions. I am working on this simple problem:

Show that
$$\overline{A\cup B}=\overline{A} \cup \overline{B}$$
where $\overline{X}$ denotes the closure of $X$.

Now I have already proved it by showing that LHS is a subset of RHS and vice-versa and using the fact that a finite union or intersection of closed sets is closed. I'm wondering if it can't be solved in an even more direct way as follows:
Let $L_{X}$ denote the set of all limit points of the set $X$. Then $\overline{A\cup B}=(A\cup B)\cup L_{A\cup B}$. $L_{A\cup B}=L_{A}\cup L_{B}$, so $\overline{A\cup B}=(A\cup B)\cup (L_{A}\cup L_{B})$. Now can I write this expression as $(A\cup L_{A})\cup(B\cup L_{B})$ so that it becomes $\overline{A}\cup\overline{B}$, which is what I need?
Apologies if this seems like a trivial question.
 A: Union is indeed a commutative and associative operation on sets (well known from set theory, we use set theory all the time in topology), and denoting the set of limit points of any set $C$ by $C'$, as is more common, you can indeed say
$$\overline{A \cup B} = (A \cup B) \cup (A\cup B)' = (A \cup B) \cup (A'\cup B')= (A \cup A') \cup (B \cup B)' = \overline{A} \cup \overline{B}$$
provided you indeed know that limit points sets are preserved by finite unions.
An alternative proof using not limt points but the definition of a $\overline{C}$ as the smallest (by inclusion) closed superset of $C$:
$\overline{A} \cup \overline{B}$ is closed (as finite sets are closed under finite unions which follows (by de Morgan) from the fact that finite intersections of open sets are open). It contains $A \cup B$ trivially, so by minimality $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$. OTOH: $A \subseteq A \cup B$ so $\overline{A} \subseteq \overline{A \cup B}$ and also $B \subseteq A \cup B$ so $\overline{B} \subseteq \overline{A \cup B}$, and combine these to give $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$ and we have both inclusions and equality.
