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.

Here's a second hint:  Suppose you have two sets with different definitions, as you do here.  Let's call them $X$ and $Y$.  Then to show that $X=Y$, it is enough to show $X\subseteq Y$ and $Y\subseteq X$.  To show $X\subseteq Y$ you take an arbitrary point $p$ in $X$ and show that $p$ is also a point of $Y$; to show $Y\subseteq X$ you do the reverse.
In this case you want to show $$\bigcap_{\substack{B\subseteq C\\\text{$C$ closed}}} C = B.$$ Let's call the thing on the left $X$ for short.  You then want to show $X=B$.  
To do this, it is enough to show that $X\subseteq B$ and that $B\subseteq X$.  Start with $X \subseteq B$.  Suppose some point $p$ is in $X$.  What properties must $p$ have?  Can you use those properties to show that $p$ must also be in $B$?
