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 $X$.
Best Answer
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