Kuratowski's definition of a topological space, and immediate consequences According to Kuratowski, a topological space is a set $X$ together with an operation ${\bf C}$ (called closure) which associates to every element $A\subset X$ an element ${\bf C}A\subset X$ such that:
(1) ${\bf C}(A\cup B)={\bf C}A\cup{\bf C} B$
(2) $A\subset{\bf C} A$
(3) ${\bf CC}A={\bf C}A$
(4) ${\bf C}0=0$
Several properties can be derived from these axioms. In particular:
(1) ${\bf C} (A\cap B)\subset {\bf C} A\cap{\bf C} B$
(2) ${\bf C}(\bigcap A_t)\subset\bigcap({\bf C} A_t)$
(3) $\bigcup({\bf C} A_t)\subset {\bf C}(\bigcup A_t)$
I have no problem with properties (1) and (2).
Concerning (3), the proof goes as follows:
For each index $m$, we have $A_m\subset\bigcup A_t$. Hence, ${\bf C}A_m\subset{\bf C}(\bigcup A_t)$, that is to say, $\bigcup({\bf C} A_t)\subset {\bf C}(\bigcup A_t)$ for $m$ was arbitrary.
What I do not understand is why the very same argument does not apply to the closure of the union of two sets. Put differently, why, in the definition of a topological space, we have ${\bf C}(A\cup B)={\bf C}A\cup{\bf C} B$, and not ${\bf C}A\cup{\bf C} B\subset{\bf C}(A\cup B)$?
 A: It's an axiom (or part of the definition) that ${\bf C}(A \cup B) = {\bf C}A \cup {\bf C}B$. This axiom implies the lemma that whenever $A \subset B$ we have ${\bf C}A \subset {\bf C}B$ (which is used in your proof of (3), and is not an axiom so needs a proof!): $A \subset B$ iff $A \cup B = B$ so ${\bf C}B = {\bf C}(A \cup B) = {\bf C}A \cup {\bf C}B$, which implies that ${\bf C}A \subset {\bf C}B$. If we had only a $\subset$ instead of $=$ in the last line, we could not have concluded that we had the final inclusion, but we'd have had a triviality.
So the axiom (1), for unions of two sets, gives us the preservation of the subset relation, and this in turn (!) gives us a subset relation for arbitrary unions, in property (1). Of course we already know that for finite unions we have equality, whereas for infinite unions we can only state the inclusion (all singleton sets of rational numbers are closed, but their union is all rationals, which is not closed; all in the reals in their usual topology).
Another way to put it: the proof for property (1), for all unions, implicitly already uses the equality for unions of two sets.    
