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?
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.