Equivalence of Definitions of Countably Compact Space

Theorem
Let $X$ be a topological space.

The following statements are equivalent:
 * $(1): \quad$ $X$ is countably compact, i.e. every countable open cover of $X$ has a finite subcover.


 * $(2): \quad$ Every countable set of closed sets whose intersection is empty has a finite subset whose intersection is empty. That is, $X$ satisfies the Countable Finite Intersection Axiom.


 * $(3): \quad$ Every infinite subset of $X$ has an $\omega$-accumulation point in $X$.


 * $(4): \quad$ Every infinite sequence in $X$ has an accumulation point in $X$.

$(3) \iff (4)$
By definition, a point is an $\omega$-accumulation point iff it is an accumulation point of that set viewed as an infinite sequence.

$(1) \iff (3)$
Suppose $X$ has a countable open cover $\left\{{U_i}\right\}$ which does not have a finite subcover.

Thus we can find a sequence $\left \langle {x_n} \right \rangle$ of distinct points such that $\displaystyle x_n \notin \bigcap_{i+1}^n U_i$.

Now every point in $X$ has a neighborhood, i.e. one of the $U_i$ to which it belongs.

But $U_i$ intersects only finitely many points of $X$.

So $\left \langle {x_n} \right \rangle$ sequence can have no $\omega$-accumulation point in $X$.

Now suppose that $S \subseteq X$ is a countably infinite subset of $X$ which does not have an $\omega$-accumulation point.

Then each $x \in X$ would have an open neighborhood $U_x$ which intersects at most finitely many points of $S$.

For each finite subset $F$ of $S$, we define $\displaystyle U_F = \bigcup \left\{{U_x: U_x \cap S = F}\right\}$.

Then $\left\{{U_F}\right\}$ is a countable open cover of $X$ such that every finite subset of it includes at most finitely many points of $S$.

Thus no finite subcollection can cover $X$.