Compact Sets in Countable Complement Space

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

Then the compact sets of $T$ are exactly the finite subsets of $S$.

Proof
Any finite set is compact.

Suppose $H \subseteq S$ is an infinite compact set.

Take a (countably) infinite sequence $(a_n)_{n\ge 0}$ of distinct elements of H.

Consider the open sets $V_m:=S\setminus \{a_{m+n}\}_{n\ge 0}$ for $m\ge 0$, which satisfy $V_{m_1}\subset V_{m_2}$ if $m_1<m_2$.

Then

$H\subseteq \displaystyle \bigcup_{m\ge 0} V_m$

is an open cover of $H$.

Since H is compact, there is a finite open subcover, say

$H\subseteq \displaystyle \bigcup_{i=0}^N V_{m_i}$

with $m_0<m_1<...<m_N$. But then

$\displaystyle \bigcup_{i=0}^N V_{m_i}=V_{m_N}$

and it follows that

$a_{m_N+1}, a_{m_N+2},... \notin V_{m_N}=\displaystyle \bigcup_{i=0}^N V_{m_i}$

which implies $a_{m_N+1}, a_{m_N+2},... \notin H$, contradiction.