Fortissimo Space is not Sequentially Compact

Theorem
Let $T = \left({S, \tau_p}\right)$ be a Fortissimo space.

Then $T$ is not sequentially compact.

Proof
Let $\left \langle {x_n}\right \rangle_{n \in \N}$ be any sequence such that:
 * $\forall n \in \N: x_n \ne p$

and that:
 * $n \ne m \implies x_n \ne x_m$

Then:
 * $\left\{{x_n: n \in \N}\right\} \subset X \setminus \left\{{p}\right\}$

By the definition of Fortissimo space, any $A \subset X \setminus \left\{{p}\right\}$ is open, because $p \notin A$.

Thus $X \setminus \left\{{p}\right\}$ as a subspace of the Fortissimo space is discrete.

From:
 * All Points in Discrete Space are Isolated

and:
 * Point is Isolated Iff Not a Limit Point

it follows that if $\left \langle {x_n}\right \rangle_{n \in \N}$ converges, then it converges to $x_k$ for some $k \in \N$.

We also know that $\left\{{x_k}\right\}$ is open in $X \setminus \left\{{p}\right\}$. So from the definition of limit, $\exists N\in\N:\forall n>N$, $x_n\in\left\{{x_k}\right\}\Rightarrow x_n=x_k$.

We concluded that $\left \langle {x_n}\right \rangle_{n \in \N}$ converges only if the sequence is constant from some $k \in \N$.

From the properties of the sequence $\left \langle {x_n}\right \rangle_{i \in \N}$, it cannot be constant from some $k$, so it cannot converge.

The result follows from the fact that every subsequence of $\left \langle {x_n}\right \rangle_{n \in \N}$ has the same properties as $\left \langle {x_n}\right \rangle_{n \in \N}$, namely:
 * $n \ne m \implies x_n \ne x_m$

and:
 * $\forall n \in \N: x_n \ne p$