If $X$ is $T_1$ and limit point compact ,then its countably compact [duplicate]

I tried to prove this by contradiction.

Assume $X$ is not countable compact, then the countable open cover $\{\cup_{i=1}^{\infty} U_i\}$ doesn't have a finite cover. So it exists a $x\in X\setminus \{\cup_{i=1}^{n} U_i\} $.

Consider the infinite set $A$ that has a limit point, the point $x$.

Then for every open n-d $(U_x)$ of $x,\:$ $U_x\cap A\setminus \{x\}\neq \varnothing $,

I don't see how I can use that $X$ is $T_1$ can someone help continue my proof ?

Hint: show that a $T_1$ space $X$ that is limit point compact is in fact even strongly limit point compact, I.e. every infinite subset $A$ has an $\omega$-limit point $p \in X$ ( every neighbourhood of $p$ contains infinitely many points of $A$).

Then showing the countable compactness will be a lot easier with your idea.

$T_1$ implies that no finite set has a limit point, and this is useful to know.