Show that every open set A in a metric space (X, d) is the union of closed sets. This question is confusing to me because I do not know how the definition of an open set pertains to the proof of the question. Because to me it seems trivial.
I can simply say that in any metric space, I can view singleton sets as closed sets and thus, the set A can be written as the union of these singletons. To me, that seems it would simply end the proof right there.
Is there another type of approach where I actually use the definition of open set? Been looking at this for a while, just not able to see it. 
 A: Your proof is perfectly fine. However, here's a proof that every open set is an union of at most countably many closed sets:
Be $U$ an open set.
First, let's consider the case $\partial U \ne \emptyset$. Then for any $n\in\mathbb N$ we define $D_n$ to be the set of points whose distance to $\partial U$ is less than $1/n$. Note that $D_n$ is an open set (it's the union of the open balls with radius $1/n$ around all points of $\partial U$).
Next we define $C_n = U\setminus D_n$. Now $C_n$ is a closed set, because its complement is open (it's the union of the open set $D_n$ and the open exterior of $U$). By definition, $C_n\subset U$.
Now $\bigcup_{n\in N} C_n = U$. This is because for any $x\in U$, $d(x,\partial U)>0$ (because $U$ is open), therefore there exists an $n\in\mathbb N$ such that $d(x,\partial U)>1/n$, and thus $x\notin D_n$. Therefore $x\in U\setminus D_n = C_n$.
Thus the $C_n$ form a family of closed sets whose union is $U$.
Remains the case $\partial U = \emptyset$. But in that case, $\partial U\subseteq U$ and thus $U$ is closed, and thus trivially the union of closed sets (namely the union of itself).
