If is an open subset in then is open in . So, if is closed in then is closed in . Now you define the closure of in as the intersection of all closed subsets of such that they contain , but since closed sets in are closed in then the closure in has more elements to intersect and thus is contained in the closure in .
Well, that is an overview of the proof. I leave the details to you.
Here is a different approach.
if and only if each for each open set , .