It's written in Willard (15G.3) that the only convergent sequences in a Hausdorff Extremally Disconnected space are the eventually constant sequences. However, it has not provided a proof.
I've tried to derive this myself, but am unable to do so. There's another post on Math Stackexchange about this problem, but a solution wasn't presented there. So, any help in proving this is appreciated!
Best Answer
Let $(x_n)$ be a sequence in $X$ such that $x_n \to p \in X$.
Assume it is not eventually constant. Then in particular $x_n \ne p$ for infinitely many $n$, i.e. there exists a subsequence $(x_{n _k})$ such that $y_k = x_{n _k} \ne p$ for all $k$. Let us inductively construct a subsequence $(y_{k_r})$ of $(y_k)$ and a sequence of pairwise disjoint open sets $U_r$ such that $y_{k_r} \in U_r$ and $p \notin \overline U_r$.
For $r=1$ we take $k_1 = 1$. There exist disjoint open neigborhoods $U_1$ of $y_1$ and $V_1$ of $p$. Thus $p \notin \overline U_1$. For the induction step observe that $W_r = X \setminus \bigcup_{i=1}^r \overline U_i$ is an open neigborhood of $p$. Since $y_k \to p$, we find $k_{r+1}$ such that $y_{k_{r+1}} \in W$. There exist disjoint open subsets $U_{r+1}$ and $V_{r+1}$ of $W_r$ such that $y_{k_{r+1}} \in U_{r+1}$ and $p \in V_{r+1}$. These subsets are also open in $X$. Clearly $p \notin \overline U_{r+1}$ and $U_{r+1} \cap \bigcup_{i=1}^r \overline U_i = \emptyset$ which shows that $U_1,\ldots, U_{r+1}$ are pairwise disjoint.
Let $U = \bigcup_{i=1}^\infty U_{2i}$ which is open. Thus also $\overline U$ is open. Since $y_{k_{2i}} \in U$ and $y_{k_{2i}} \to p$, we have $p \in \overline U$. Therefore $y_{k_r} \in \overline U$ for $r \ge R$. Now let $r \ge R$ be odd. Then $U_r \cap U = \emptyset$, thus $U_r \cap \overline U = \emptyset$ because $U \subset X \setminus U_r$. We conclude that $y_{k_r} \notin \overline U$. This is the desired contradiction.