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

We have that $\left \langle {x_n}\right \rangle_{n \in \N}$ is infinite.

Suppose $\left \langle {x_n}\right \rangle$ also takes infinite different values in $S$.

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

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

Then $X \setminus U$ 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 different values.

Then, since the sequence is infinite, there exists $x \in S$ such that $\forall N \in \N: \exists n > N$ so that $x = x_n$.

This implies that we can take a subsequence of the original sequence which is constant, and then converges to that constant.

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