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 $p \in \cl{E} \cap \cl{O}$. Yet by definition, $E \cap O = \emptyset$, so the space is not extremally disconnected.

Let $A \subseteq X$ such that, for any $K$ compact $T_2$ and $f : K \to X$ continuous, then $f^{-1}(A)$ is closed. We wish to show $A$ is closed. To do that, we take a sequence $(x_n)$ in A such that $x_n \to x$ and show $x \in A$ (using that $X$ is sequential).

We use a nice trick: $\omega + 1$ is compact and $T_2$ under the order topology. If we define $f : \omega + 1 \to X$ as $f(n) = x_n$ and $f(\omega) = x$, then $f$ is continuous: Note that $\omega \subseteq \omega + 1$ is discrete, so clearly it is continuous at each $n < \omega$, and for any nbd $U$ of $x$, there is some $n_0$ such that $x_n \in U$ for $n > n_0$, so $f((n_0, \omega]) \subseteq U$ and $f$ is continuous at $\omega$. This proves $f^{-1}(A)$ is closed. $x \notin A$ would imply $f^{-1}(A) = \omega$, which is not a closed set ($\omega \in \cl{\omega}$).

Let $L$ be compact. Using the $k_2$-space property, we wish to show that for any $K$ compact $T_2$ and any $f : K \to X$ continuous, that $f^{-1}(L)$ is closed. Our objectives are to show (1) $X$ is $T_1$ (2) $f(K)$ is compact $T_2$ (3) prove $f^{-1}(L)$ is closed.

(1) The inclusion map $\text{id} : \{x\} \to X$ is continuous and $\{x\}$ is clearly compact $T_2$, so apply Weak Hausdorff.

(2) $f(K)$ is compact since $f$ is continuous. Note that $T_2$ spaces separate compact sets, and since every closed set in $K$ is compact, $K$ is $T_4$. Take $x \ne y \in f(K)$. $\{x\}$ and $\{y\}$ are closed in $X$ by (1), so $A = f^{-1}(\{x\})$ and $B = f^{-1}(\{y\})$ are closed and can be separated by open nbds $A \subseteq U$ and $B \subseteq V$. Note that $K \setminus U$ is compact, so $f(K \setminus U)$ is closed and $U' = f(K) \setminus f(K \setminus U)$ is a nbd of $x$. Similarly, $V' = f(K) \setminus f(K \setminus V)$ is a nbd of $y$ and $U' \cap V' = \emptyset$.

(3) $L' = f(K) \cap L$ is compact. Since $f(K)$ is $T_2$, $L'$ is closed in $f(K)$ But $f(K)$ is closed in $X$ by Weak Hausdorff, so $L'$ is closed. Then clearly $f^{-1}(L) = f^{-1}(L')$ is closed.

By definition, let $X$ be a countable union of increasing compact sets $\{K_n\}$. Now define the sequence of sets

$$L_n = \{ x \in X \ : \ \cl{\{x\}} \cap K_n \ne \emptyset \}$$

Clearly by construction, $K_n \subseteq L_n$ for each $n$, $\{L_n\}$ is increasing, and $X = \bigcup_{n < \omega} L_n$. Now let $\mathcal{U}$ be an open cover of $L_n$. Since it also covers $K_n$, let $\{U_1, \dots, U_m\}$ be a countable subcover of $K_n$. For any $x \in L_n$, there is some $y \in \cl{\{x\}} \cap K_n$. Thus, $y \in U_j$ for some $1 \le j \le m$, but that implies $x \in U_j$. So this is also a finite subcover of $L_n$, proving each $L_n$ is compact.

By contraposition, suppose $K$ is not contained in any $L_n$. So choose $x_n \in X \setminus L_n$ for each $n$, so that $\cl{\{x_n\}} \cap K_n = \emptyset$. Then define $V_n = X \setminus \bigcup_{k=n}^\infty \cl{\{x_k\}}$. We see that $\cl{\{x_k\}}$ will only intersect $K_m$ for $k < m$, so that $V_n \cap K_m = K_m \setminus \bigcup_{k=n}^{m-1} \cl{\{x_k\}}$, which is open in $K_m$. By the $k_{\omega,1}$-space property, $\{V_n\}$ is an open cover of $K$: for any $x$, $x \in K_n$ for some $n$ and so $x \notin \cl{\{x_n\}} \implies x \in V_n$. Note that it is also increasing, which means if it had a finite subcover, we’d have $K \subseteq V_n$ for some $n$, which would imply $x_n \notin K$. So $K$ is not compact.

Source: M W on stack exchange, awesome trick