Prove that a T1 space that is countably compact is limit point compact I was wondering if someone could review my proof for correctness. Thanks! 
Proof:
Let X be a topological space. (don't need T1 for this direction) 
Let X be  a countably compact space. 
Suppose X is not limit point compact. 
Then take some infinite subset A that has no limit points. 
Suppose A is countably infinite. 
Then A is closed since $\bar{A}$ = A. 
If A contains no limit points we can take open sets around each a $\in$ A such that for the open set U$_a$ containing a $\in$ A we have that U$_a$ intersects A in no points other than $a$ itself. The union: 
$\bigcup$ U$_a$ for each a $\in$ A covers all of A. 
The above union along with X - A is an open covering of X. It is countable since A is countable, in this case. Hence we can find a finite subcollection and this requires us to reduce the infinite collection of $\bigcup$ U$_a$ to a finite union. But since each U$_a$ only contains one point of A, any finite subcollection will not cover A, since A is infinite. This inability to reduce a countable open covering to a finite open covering is a contradiction since X is countably compact. 
Now suppose A was instead uncountably infinite. 
Then we can find some countably infinite subset of A. Call this set B. 
Then B cannot have any limit points either since if B did have a limit point it would also be a limit point of A (which by hypothesis has no limit points). 
Then B must be closed and so X - B is open. 
Then similarly to above we can construct an open covering of B with open sets, U$_b$, around each b $\in$ B that intersects B in no points other than b itself.
Hence we have the countable open covering of X: 
X- B $\bigcup$ ($\bigcup$U$_b$ for each b $\in$ B)  
Since X is countably compact we can again make this collection finite and still cover X. But since each U$_b$ was chosen to only contain one  distinct point of B, any finite subcollection $\bigcup$U$_b$ will not cover B. This is a contradiction since X is countably compact.
Hence X must be limit point compact. 
 A: The structure of the proof is not quite clear (to me). The uncountable/countable distinction is moot, only countable matters:
First note:

Fact 1: a countable space in the discrete topology is not countably compact.
Fact 2: a closed subset of a countably compact space is countably compact.

So we assume throughout that $X$ is countably compact (in the countable open cover sense). So we have to prove that every infinite subset $A$ of $X$ has a limit point. So let $N$ be a countably infinite subset of $A$ (which exists, as $A$ is infinite, minor use of the Axiom of Choice here). I claim that $N$ already has a limit point: for suppose not, then $N$ and all its subsets have no limit points (and a set with no limit points is closed) and this means that $N$ (as a subspace of $X$) is closed and discrete, and this combination gives us a contradiction: $N$ would contradict fact 1 because it is countably compact by fact 2. So $N$ does have a limit points, and thus so does the larger set $A$. QED.
As an alternative argument for $N$, we note $N$ is closed when it has no limit points and pick $U_x$ open such that $U_x \cap N = \{x\}$ for all $x \in N$ by the fact that $x \in N$ is no limit point of $N$, and then we can use the countable cover $X\setminus N$, $\{U_x: x \in N\}$ of $X$ directly. This is also the OP's argument. It comes down to the same thing, really.
So the original proof by the OP was in essence correct, but does not need a separate uncountable case: once you have shown all countably infinite subsets have a limit point, all infinite subsets have a limit point, and you're done.
Note that this direction indeed does not $T_1$-ness. The reverse does, however.
