Countably compact means : every countable open covering contains a finite subcollection that covers it.
Limit point compact means: every infinite set contained in it has a limit point.
In T1 space every singleton is closed and if x is a limit point of some subset A, then every neighborhood of x contains infinitely many points of A.
Can anyone prove :
If X is limit-point compact space,which is T1,then X is countably compact.
Let $\{U_n\}_{n}$ be a countable cover of $X$ and assume that this cover does not have a finite subcover. Then for every $n \in \mathbb{Z}_{>0}$ there exists $x_n\in X\setminus\bigcup_{k = 1}^n U_k$ and we can set $A = \{x_n \:|\: n \in \mathbb{Z}_{>0}\}$. This is an infinite set because every element of $X$ is contained in some $U_n$ and there are only finitely many $k$ such that $x_k \in U_n$ (because $x_k \notin U_n$ for $k \geq n$). This also shows that $A \cap U_n$ is finite for all $n$. By limit-point compactness we get a limit point $x \in X$ of $A$. This is contained in $U_n$ for some $n$. But then $U_n$ contains infinitely many elements of $A$ in contradiction to $A \cap U_n$ being finite.
