Compact $\Rightarrow$ countably compact: Every countable open cover is an open cover, so it has a finite subcover.

Countably compact $\Rightarrow$ limit point compact: Suppose $A$ is an infinite subset with no limit point. Then $A$ is closed (its complement is open since no point outside $A$ is a limit point). For each $a \in A$, there exists an open $U_a$ with $U_a \cap A = \{a\}$ (since $a$ is not a limit point of $A \setminus \{a\}$). Choose a countably infinite subset $\{a_n\} \subseteq A$. Then $\{U_{a_n}\} \cup \{X \setminus A\}$ is a countable open cover of $X$ with no finite subcover (each $a_n$ is only covered by $U_{a_n}$).

Sequentially compact $\Rightarrow$ countably compact: Suppose $X$ is sequentially compact and $\{U_n\}_{n \geq 1}$ is a countable open cover with no finite subcover. For each $n$, choose $x_n \in X \setminus (U_1 \cup \cdots \cup U_n)$. By sequential compactness, $(x_n)$ has a subsequence converging to some $x \in X$. Since $\{U_n\}$ covers $X$, $x \in U_m$ for some $m$. But $x_n \notin U_m$ for $n \geq m$, so no subsequence of $(x_n)_{n \geq m}$ can converge to $x$ unless the subsequence eventually enters $U_m$ -- which it does since $U_m$ is a neighborhood of $x$. But $x_{n_k} \notin U_m$ for $n_k \geq m$ and $x_{n_k} \to x \in U_m$, so eventually $x_{n_k} \in U_m$, contradiction.

We sketch the key implications not already proved.

(3 $\Rightarrow$ 4 in metric spaces): Let $(x_n)$ be a sequence. If the range $\{x_n\}$ is finite, some value repeats infinitely often, giving a constant (hence convergent) subsequence. If the range is infinite, it has a limit point $x$ by (3). For each $k$, the ball $B(x, 1/k)$ contains infinitely many $x_n$. Inductively choose a subsequence $x_{n_k} \in B(x, 1/k)$; then $x_{n_k} \to x$.

(4 $\Rightarrow$ 1 in metric spaces): This is the most involved step and uses total boundedness. First, sequential compactness implies total boundedness: if not, there is an $\epsilon > 0$ and a sequence with $d(x_m, x_n) \geq \epsilon$ for all $m \neq n$, having no convergent subsequence. Second, use the Lebesgue covering lemma (which holds in sequentially compact metric spaces) to convert any open cover to a finite subcover.