Countable Complement Space is not Countably Compact

Theorem
Let $T = \left({S, \tau}\right)$ be a countable complement topology on an uncountable set $S$.

Then $T$ is not a countably compact space.

Proof
Consider $U \in \tau$.

By definition of countable complement topology, $\complement_S(U)$ is countably infinite.

Then for each $x\in \complement_S(U)$, $\complement_S(U)\setminus \{x\}$ is countably infinite.

Then for each $x\in \complement_S(U)$, $\complement_S(\complement_S(U)\setminus \{x\}) = U \cup\{x\}$ is open.

Hence $\{U\cup \{x\}:\ x\in\complement_S(U)\}$ is an open cover.

This cover is countable because is equivalent to $\complement_S(U)$.

Suppose there exists a finite subcover $\{U \cup \{x_1\}, \ldots, U \cup \{x_n\}\}$ of $T$.

Then:
 * $(U\cup\{x_1\})\cup\ldots\cup(U\cup\{x_n\}) = U \cup\{x_1,\ldots,x_n\}$

Since $\complement_S(U)$ is not finite, it follows that $U \cup\{x_1, \ldots, x_n\} \ne S$.

Thus there is no finite subcover of $T$.