Fortissimo Space is not First-Countable

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

Then $T$ is not a first-countable space.

Proof
This proof follows the proof from Countable Complement Space is Not First Countable.

Suppose that $p \in S$ has a countable local basis.

That means there exists a countable set of sets $\mathcal B_p \subseteq \tau$ such that $\forall B \in \mathcal B_p: p \in B$ and such that every open neighborhood of $p$ contains some $B \in \mathcal B_p$.

So:

By definition of the fortissimo space, each of $S \setminus B$ is countable.

From Union of Countable Sets it follows that $\displaystyle \bigcup_{B \in \mathcal B_p} \left({S \setminus B}\right)$ is also countable.

So $S \setminus \left\{ {p}\right\}$ and therefore $S$ is also countable.

From this contradiction (as we have specified that $S$ is uncountable) it follows that our assumption that $p \in S$ has a countable local basis must be false.

Hence by definition $T$ can not be first-countable.