If the closure of a set $A$ is defined as the intersection of all closed sets which contain $A$, prove that closure of a closed set $B$ is $B$ itself If the closure of a set $A$ is defined as the intersection of all closed sets which contain $A$, prove that closure of a closed set $B$ is $B$ itself.
Attempt: I apologize if this is too basic but I am taking an inttoductory course in Elements of Real Analysis.
Let $B$ be a closed set, then closure of a $B$ is defined as the intersection of all closed sets which contain $B$.
Hence, we wants closed sets $B_1,B_2, \cdots , B_m$ such that $B_1 \bigcap B_2 \bigcap \cdots \bigcap B_m = B$
Such that $B \subseteq B_1,B \subseteq B_2,~~\cdots~~, B \subseteq B_n$
We need to prove using the given definition that $B_1 \bigcap B_2 \bigcap \cdots \bigcap B_m = B$
Since, intersection of closed sets is a closed set $\implies B_1 \bigcap B_2 \bigcap \cdots \bigcap B_m = A$ where $A$ is a closed set and $B \subseteq A$ ( Where $B$ is also a closed set )
How do I move ahead using the given definition. I am aware of the method of solving the problem using the derived set method, but the problem seeks to solve it through the given definition only.
EDIT : Since, one of the closed sets that contains B is B itself, then is it appropriate to take $B_1=B_2=⋯B_m=B $. But, then what about the other closed sets which contain $B$? 
Thank you for your help.

This is merely a question of logic. It has not much to do with the exact definition of closedness, nor with finite vs. infinite intersections, etc.
If $B$ is a closed set and $B$ its closure according to definition then
$$B\subset\bar B\subset B\ ,\tag{1}$$
and this implies $B=\bar B$. In $(1)$ the first inclusion holds because all sets $A$ entering the intersection $\bar B$ contain $B$, and the second inclusion holds because $B$ itself is one of these sets $A$.
