Fort Space is Sequentially Compact

Theorem
Let $T = \left({S, \tau_p}\right)$ be a Fort space on an infinite set $S$.

Then $T$ is a sequentially compact space.

Proof
Let $\left \langle {x_n}\right \rangle_{n \mathop \in \N}$ be an infinite sequence in $T$.

Suppose $\left \langle {x_n}\right \rangle$ takes an infinite number of distinct values in $S$.

Then there is an infinite subsequence $\left \langle {x_{n_r}}\right \rangle_{r \mathop \in \N}$ with distinct terms.

Let $U$ be a neighborhood of $p$.

Then $S \setminus U$ is a finite set by definition.

Thus there exists $N \in \N$ such that $\forall r > N: x_{n_r} \in U$.

Thus $\left \langle {x_{n_r}}\right \rangle$ converges to $p$.

Otherwise $\left \langle {x_n}\right \rangle$ only takes a finite number of distinct values.

Then, since $\left \langle {x_n}\right \rangle$ is infinite, there exists $x \in S$ such that:
 * $\forall N \in \N: \exists n > N: x = x_n$

This implies that we can take a subsequence of $\left \langle {x_n}\right \rangle$ which is constant, and which converges to that constant.

We can conclude then that, by definition, $T$ is a sequentially compact space.