Proving, that closure of set is equal this set iff set is closed

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I've started intorduction to topology course and I need help with one of the problems:

Let $A \subset(X,T). $ Prove that $cl(A) = A\iff A$ is closed.

It may looks trivial, but I had a little (too long...) break from math and I'm struugling now. Need some help and time to understand problems.
Also, if anyone can recommend good book/website with introduction to topology and/or functional analysis, I'd be grateful :)
Have a nice math day!

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The closure $\operatorname{cl}(A)$ of a subset $A$ of $X$ is defined to be the intersection of all closed sets in $X$ containing $A$. Since the arbitrary intersection of closed sets is closed, it follows that the closure $\operatorname{cl}(A)$ is closed in $X$.

Now suppose that $A=\operatorname{cl}(A)$. Since $\operatorname{cl}(A)$ is closed, it follows that $A$ is closed.

Conversely, if $A$ is closed, then $A$ is in the intersection of all closed sets containing $A$, and is necessarily the smallest such set (ordered by inclusion). Thus, the intersection will be equal to $A$, and we find that $A=\operatorname{cl}(A)$.

As far as good topology introductions, Munkres' Topology is an indispensable resource.