Do all the open sets containing a limit point of an infinite countably compact subset have to contain infinite points? Say an infinite set is countably compact (if set $E$ is an infinite countably compact set, it contains at least one limit point within itself). Let $x$ be one such limit point of $E$. My textbook says "if this is to be a compact space, every open set containing $x$ has to contain an infinite number of points from $E$". 
Is this a necessary condition? Say the base $\mathfrak{B}$ of the topology contains an open set which contains all but a finite number of points from $E$. Let this set be $A$. $A$ may intersect with other sets of $\mathfrak{B}$. However, each of these intersections has to contain all but finite points $\in A$. If these conditions are satisfied, and $E$ is the ony infinite subset of $X$, then $X$ has a finite subcover for every cover, and is hence a compact space! 
$x$ may be contained in just two open sets- $X$ and an open set containing a finite number of points from $E$ along with $x$. This makes $x$ a limit point of the infinite set $E$, makes the infinite set $E$ countably compact by giving it a limit point within itself, and also makes $X$ compact if $E$ is the only infinite subset of $X$. 
 A: Something may be wrong:
In fact, Assuming $X$ is $T_1$, if $x$ is a limit of $E$, for any open set $U$ of $x$ contain infinite number of points from $E$.  So $E$ don't need to be compact.
Example: Let $X=[0,\omega_1)$. It is a countably compact space, since every limit point is in $X$. However it is not compact. Note that for every limit point $x$ of $X$, its every open nbhd contains infinite pints from $X$.

Here is something related your question which you may be interested in:
Let $X$ be a space and $A$ a subset of $X$. A point $x\in X$ is a point of complete accumulation of $A$ if $|U\cap A|=|A|$ for every open nbhd $U$ of $x$. 
A space $X$ is compact iff every infinite set in $X$ has a point of complete accumulation. 
Proof: see Here
A: There is are different definitions, for a general space $X$:


*

*Every infinite subset $E$ of $X$ has a limit point.

*Every infinite subset $E$ of $X$ has a $\omega$-limit point.


Here $x$ is a limit point of $E$ when every open set $O$ that contains $x$ contains a point from $E$ that is different from $x$.
And $x$ is an $\omega$-limit point of $E$ when every open set $O$ that contains $x$ contains infinitely many points of $E$.
Clearly the second definition implies the first, because if $O$ contains infinitely many points of $E$, one of them is surely different from $x$. In a $T_1$ space (so where finite sets are closed) the first does imply the second, as is easily checked. And e.g. metric spaces are $T_1$. 
An example of a space that satisfies the first definition (sometimes called limit point compact) but not the second, is a countable discrete space (say $\mathbb{N}$) times the indiscrete 2 point set $\{0,1\}$ (the so-called doubling of the space) in the product topology. Then if a(n infinite) set $E$ contains, say, $(n,0)$ then $(n,1)$ is a limit point of $E$, as a basic neighbourhood of that point is $\{(n,0), (n,1)\}$ which contains a point of $E$ that is unequal to itself. 
The second definition is called countably compact, and is equivalent to "every countable open cover of $X$ has a finite subcover" (see this answer, e.g.). In metric spaces it's equivalent to compactness, but in general it's not.
A compact space always does satisfy the second (and so the first) definition. 
