Closed property of nonempty finite set

Disregarding possible fine print of definitions the resolution is quite simple:

If there are no limit points, then of course all limit points are trivially contained in the set.

"$A$ contains all of its limit points" is short for "for all $x$ such that $x$ is a limit point of $A$, we have that $x \in A$".

This statement is voidly true if there are no limit points (as in this case).

Equivalently put, the condition can be written as $A' \subseteq A$ and the empty set is a subset of all sets.

Alternatively a set $A$ is closed iff $A=\bar A= A \cup \{\text{set of all limit points}\}.$

In your case, $\{\text{set of all limit points}\}=\emptyset.$ Thus $A=\bar A$.