Why is $\operatorname{cl}(A)$ the smallest closed set containing $A$? If $\operatorname{cl}(A)$ is the intersection of all closed subsets of $X$ containing $A$.
How to prove that $\operatorname{cl}(A)$ is the smallest closed set $C$ in $X$ such that $A ⊆ C$?
I have proved $\operatorname{cl}(A)$ is closed, but then I'm stuck for hours. 
 A: What we mean when we say that $C=\operatorname{cl}(A)$ is the "smallest" closed set such that $A\subseteq C$ is that if, for any other closed set $D$ with $A\subseteq D$, it is true that $C\subseteq D$. Since $C$ is contained within any closed set containing $A$, it is the "smallest" in this sense.
Let's prove this is true. Suppose that $D$ is a closed subset of $X$ with $A\subseteq D$. Recall that, by the definition of $\operatorname{cl}(A)$ that you're given, 
$$\operatorname{cl}(A)=\bigcap_{\substack{\text{closed }E\;\subseteq X\\ A\,\subseteq E}} E.$$
But $D$ is one of those closed sets that "$E$" ranged over in the intersection - i.e., 
$$\operatorname{cl}(A)=D\cap \bigg(\bigcap_{\substack{\text{closed }E\;\subseteq X\\ A\,\subseteq E\\ E\neq D}} E\bigg).$$
For any sets $Y$ and $Z$, it is true that $Y\cap Z\subseteq Z$. In other words, if I start with a set $Z$, and intersect it with any other set, all I can do is make it smaller (at most, it will stay the same). Thus, we see that $\operatorname{cl}(A)\subseteq D$, since $\operatorname{cl}(A)$ is equal to $D$ intersected with some other sets.
Thus, we have shown that for any closed $D\subseteq X$ with $A\subseteq D$, we have that $\operatorname{cl}(A)\subseteq D$ - which is precisely what it means for $\operatorname{cl}(A)$ to be the smallest closed set containing $A$.
A: It's essentially by the definition. If there were some smaller closed set $B$ such that
$$
A \subset B \subset cl(A)
$$
with $B \subset cl(A)$ strict, then the intersection $cl(A) \cap B$ would be strictly smaller than $cl(A)$. But this is impossible, since we define $cl(A)$ to be the intersection over all closed sets containing $A$, so any such $B$ was already included in the intersection. 
