Countable Finite Complement Space is not Path-Connected

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

Then $T$ is not path-connected.

Proof
$f: \left[{0 \,.\,.\, 1}\right] \to S$ is a path on $T$.

Then $f$ is by definition continuous.

By definition of the finite complement topology, for all $x \in S$, the set $\left\{{x}\right\}$ is closed.

Now consider the set:
 * $F = \left\{{f^{-1} \left({x}\right): x \in S}\right\}$

From Continuity Defined from Closed Sets, each of the elements of $F$ is closed.

Also, from Mapping Induces Partition on Domain, the elements of $F$ are pairwise disjoint.

As $T$ is a countable finite complement topology, its underlying set $S$ is countably infinite by definition.

Hence $F$ is also countably infinite by nature of $f$ being a mapping.

Furthermore, we have $\displaystyle \bigcup F = \left[{0 \,.\,.\, 1}\right]$.

Hence $F$ is a countably infinite set of pairwise disjoint closed sets whose union is $\left[{0 \,.\,.\, 1}\right]$.

From Closed Unit Interval is not Countably Infinite Union of Disjoint Closed Sets, this is impossible.

From this contradiction, $f$ cannot be continuous, and so cannot be a path on $T$.

So $T$ cannot be path-connected.