Proof of "Every compact space is countable compact" I was reading "Introductory Real Analysis" by Kolmogorov and Fomin and came across this theorem:

Every infinite subset $S$ of a compact space $T$ has a limit point 

Note: In this book a point $x$ is called a limit point of $E$ if every neighborhood of $x$ contains an infinite number of points of $E$
Proof: Suppose $S$ has no limit point. Then there is a countable subset $R \subset S$ that has no limit point. Let $R = \{x_1, x_2, ...\}$. But then the sets $R_n = \{x_n, x_{n+1}, ...\}$ are closed sets with the finite intersection property with empty intersection. Hence $T$ is not compact.
What I don't understand is, why are these sets closed? If we assume that $T$ is a $T_1$-space, then I was able to show that they are in fact closed. Is this true if we don't assume that $T$ is $T_1$? If yes, how does one prove this?
 A: This proof is indeed incorrect (with your definition of limit point) if the space is not $T_1$, since the sets $R_n$ need not be closed.  Here is a correct argument you can give.
Let $C_n$ be the closure of the set $R_n$, so that the sets $C_n$ are clearly closed and have the finite intersection property.  We only need to check that $\bigcap C_n$ is empty.  But if $x\in\bigcap C_n$, that means that for all $n$, every neighborhood $U$ of $x$ intersects $R_n$.  If $U$ contained only finitely many points of $R$, then $U$ would be disjoint from $R_n$ for some $n$ (just pick $n$ larger than the index of any of the points of $U\cap R$).  So this means $U\cap R$ is infinite for every neighborhood $U$ of $x$.  That is, $x$ is a limit point of $R$ and hence of $S$, which is a contradiction.
A: A set without limit points is trivially closed, since it contains all of its limit points
To elaborate on Arthur's comment, while it's true that this sets will not be closed in a space with the trivial topology that's not relevant here, since they contain plenty of converging subsequences, in fact every sequence in a trivial space converges to every point of the space.
