if $X$ is $T_1$ and limit point compact then $X$ is countably compact A space $X$ is said to be countably compact if every countable open covering of $X$ contains a finite subcollection that covers $X$. I want to show that if $X$ is a $T_1$ space and limit point compact then $X$ is countably compact.
 A: Let $(U_n)_{n \in \mathbb{N}}$ be a cover of $X$ that has no finite subcover. Then for every $n$, let $x_n$ be an element of $X$ not in $\bigcup_{i \leq n} U_i$. Let $x \in X$. We will prove that $x$ is not a limit point of $\{x_n \ \vert \ n \in \mathbb{N}\}$. 
There is $n$ such that $x \in \bigcup_{i \leq n} U_i =: U$. Let $V_1$, $V_2$,..., $V_n$ be open neighborhoods of $x$ such that for each $i \leq n$, $x_i \not \in V_i$ (such neighborhoods exist because $X$ is $T_1$). Then $W := U \cap \bigcap_{i \leq n} V_i$ is an open neighborhood that does not contain $x_i$, for every $i$. Therefore, $x$ is not a limit point of $\{x_n \ \vert \ n \in \mathbb{N}\}$.
A: Suppose that the space is limit compact and is not countably compact.
The space is not countably compact open sets $U_1,U_2,\dots$ exist with $X=\bigcup_{i=1}^{\infty}U_i$ and $X\neq\bigcup_{i=1}^{n}U_i$ for every $n\in\mathbb N$.
So the sets $F_i:=U_i^c$ are closed with $\bigcap_{i=1}^{\infty} F_i=\varnothing$ and $\bigcap_{i=1}^{n} F_i\neq\varnothing$ for each $n$.
Let $x_n\in\bigcap_{i=1}^{n} F_i$ for each $n$ and let $A:=\{x_n\mid n\in\mathbb N\}$. 
For every $n$ the set $\{k\in\mathbb N\mid x_k=x_n\}$ must be finite (if not then $x_n\in\bigcap_{i=1}^{\infty} F_i$).
Then $A$ must be an infinite set so has a limit point $x$ since the space is limit compact.
Now fix $m\in\mathbb N$.
Since the space is $T_1$ we can find a neighborhood $U$ of $x$ that has empty intersection with $\{x_1,\dots,x_{m-1}\}$ and conclude that $x$ must be a limit point of set $\{x_m,x_{m+1}\dots\}\subseteq F_m$.
This implies $x\in F_m$ and this is true for every $m$, so that $x\in\bigcap_{i=1}^{\infty} F_i$. 
A contradiction is found.
