Difficulty: 5

Contrapositively, suppose $(x_n)$ is a sequence with $x_n \to p$ yet has infinitely many terms. If needed, take an injective subsequence. If $V$ is a nbd of $p$ that is $T_2$, then there’s a $n_0$ with $x_n \in V$ for $n \ge n_0$. So by another subsequence, we can assume $(x_n)$ is an injective converging sequence in a $T_2$ space.

For each $n$, it’s possible to construct a neighborhood $V_n$ which only contains $x_n$ and no other $x_m$, nor $p$: Let $x_n \in V$ and $p \in U$ with $U \cap V = \emptyset$. $x_m \in U$ for all $m \ge n_0$, so apply the Hausdorff condition finitely many times to terms $x_m < n_0$ to construct such $V_n \subseteq V$.

Now note that $x_{2n} \to p$ and $x_{2n+1} \to p$, so if $E = \bigcup V_{2n}$ and $O = \bigcup V_{2n+1}$, then any nbd of $p$ must intersect $E$ and $O$, so $p \in \overline{E} \cap \overline{O}$. Yet by definition, $E \cap O = \emptyset$, so the space is not extremally disconnected.

Let $\mathcal{R} = \{R_j\}$ be a point-countable open refinement of some open cover $\{U_i\}$. Contrapositively, suppose no countable subcover exists. Recursively, define $x_0$ as any point and $\mathcal{R}_0$ the countable collection of sets in $\mathcal{R}$ containing $x$. Then $X \setminus \mathcal{R}_0$ is nonempty and we can choose $x_1$ from that. Inductively, let $\mathcal{R}_n \subseteq \mathcal{R}$ of the sets containing $x_n$, so that $\bigcup_{j=0}^n \mathcal{R}_j$ is countable and so $x_{n+1} \in X \setminus (\bigcup_{j=0}^n \mathcal{R}_j)$.

The sequence $Y = \{x_n\}$ is infinite and has no cluster point: If $x$ were to be, take $x \in R_j$ and the minimum $n$ such that $x_n \in R_j$. By construction, it is the only element of $Y$ in $R_j$. For any sequence with no cluster point, we can construct a countable cover with no finite subcover. Take $U_0$ nbd of $x_0$, let $m$ be the least element of which $x_m \notin U_0$ and $U_1$ a nbd of $x_m$. Then if $m$ is the least element with $x_m \notin \bigcup_{k \le n} U_k$, let $U_{n+1}$ a nbd of $x_m$. Finally, if $V = X \setminus \bigcup U_n$, then $\{V\} \cup \{U_n\}$ is a countable cover of $X$ with no finite subcover.

Given $p \in \overline{A}$, choose $D \subseteq A$ countable with $p \in \overline{D}$. Suppose $(x_\alpha)_{\alpha < \lambda}$ is some transfinite sequence in $D$ with $x_\alpha \to p$. We now construct a sequence in $D$ which converges to $p$ and we’re done (thanks to PatrickR).

If some $y \in D$ is cofinal in $(x_\alpha)$, this means any neighborhood $U$ of $x$ must contain $y$, so we can construct a constant sequence $y_n = y$ with $y_n \to p$. If no $y \in D$ is cofinal, then each $I_y = \{ \alpha < \lambda \ : \ x_\alpha = y \}$ is finite and $\lambda = \bigcup_{y \in D} I_y$ is a countable union of finite sets, so it is a countable ordinal. We can assume $\lambda$ to be a regular cardinal, so it is already a sequence.