Equivalence of Definitions of Countably Compact Space

$(1) \implies (4)$: Countably Infinite Set in Countably Compact Space has $\omega$-Accumulation Point
This is proven in Countably Infinite Set in Countably Compact Space has $\omega$-Accumulation Point.

$(4) \implies (3)$
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $S$.

Let $A \subseteq S$ be the range of $\sequence {x_n}$.

If $A$ is finite, then consider the equality:
 * $\ds \N = \bigcup_{y \mathop \in A} \set {n \in \N: x_n = y}$

Therefore, there exists a $y \in A$ such that $\set {n \in \N: x_n = y}$ is an infinite set.

Hence, $y$ is an accumulation point of $\sequence {x_n}$.

Otherwise, $A$ is countably infinite.

By assumption, $A$ has an $\omega$-accumulation point $x \in S$.

It follows that $x$ is an accumulation point of $\sequence {x_n}$.

$(3) \implies (1)$
We use a Proof by Contradiction.

$S$ has a countably infinite open cover $\set {U_n: n \in \N}$ which does not have a finite subcover.

Then, using the axiom of countable choice, we can obtain a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that:
 * $\ds \forall n \in \N: x_n \notin \bigcup_{k \mathop = 0}^n U_k$

Let $x \in S$.

Then, by the definition of a cover, there exists an $n \in \N$ such that $x \in U_n$.

By construction:
 * $U_n \subseteq \set {x_0, x_1, \ldots, x_{n - 1} }$

Hence $x$ is not an accumulation point of $\sequence {x_n}$.

$(4) \implies (5)$
Follows directly from Infinite Set has Countably Infinite Subset and the definition of an $\omega$-accumulation point.