Compact implies countably compact, Lindel"of and limit point compact; countably compact together with Lindel"of implies compact; and, at the cost of countable or dependent choice, sequentially compact implies countably compact, countably compact implies limit point compact, and the converse holds when every singleton is closed

Claim 1(a): an at most countable open cover of $X$ is in particular an open cover, so compactness gives it a finite subcover, and $X$ is countably compact; and a finite subcover of an open cover is an at most countable subcover by [L2], so $X$ is Lindelöf.

Claim 1(b): let $\mathcal{U}$ be an open cover of a countably compact Lindelöf space; Lindelöfness gives an at most countable subcover $\mathcal{V}\subseteq \mathcal{U}$, countable compactness gives a finite subfamily of $\mathcal{V}$ with union $X$, and that subfamily is a finite subfamily of $\mathcal{U}$ with union $X$.

Claim 1(c): let $X$ be compact and let $A\subseteq X$ have no limit point in $X$; then $\mathcal{U}:=\{U\in \mathcal{T}:U\cap A\text{ has at most one element}\}$ covers $X$, since each $x\in X$ has a neighbourhood $N$ with $N\cap (A\setminus \{x\})=\mathrm{\varnothing }$ and an open $U$ with $x\in U\subseteq N$, so that $U\cap A\subseteq \{x\}$; compactness gives $U_0,\dots ,U_n\in \mathcal{U}$ with $X=U_0\cup \cdots \cup U_n$, whence $A=(U_0\cap A)\cup \cdots \cup (U_n\cap A)$ is listable and so finite by [L2]. Contraposing, every infinite subset of $X$ has a limit point.

For claim 2 assume countable choice, let $X$ be sequentially compact and let $\mathcal{U}$ be an at most countable open cover of $X$ with no finite subcover; then $\mathcal{U}\mathrm{\ne }\mathrm{\varnothing }$, since the empty family covers only the empty space, so [L7] lists it as $(U_n{)}_{n\in \mathbb{N}}$, and $E_n:=X\setminus (U_0\cup \cdots \cup U_n)$ is nonempty for every $n$, so [L5] supplies a sequence $(x_n)$ with $x_n\in E_n$ for every $n$.

For claim 1(d) let $X$ be countably compact and let $B\subseteq X$ be countably infinite with no limit point in $X$; fix a bijection $k\mapsto b_k$ of $\mathbb{N}$ onto $B$ and put $C_n:=\{b_k:k\ge n\}$. Each $C_n$ is a subset of $B$, so it has no limit point either by [L3], and therefore $\overline{C_n}=C_n\cup \mathrm{\varnothing }=C_n$ and $C_n$ is closed.

For claim 3 assume dependent choice and let $A\subseteq X$ be infinite. Let $S$ be the set of injections $s:n\to A$ with $n\in \mathbb{N}$, nonempty because the empty function belongs to it, and relate $s$ to $t$ when $t$ is an injection $\sigma (n)\to A$ extending $s:n\to A$; this relation is entire on $S$, since an injection $s:n\to A$ cannot have range $A$, as that would make $A$ finite, so some $a\in A\setminus \mathrm{ran}(s)$ gives the extension $s\cup \{(n,a)\}$. By [L6] there is a sequence $(s_k)$ in $S$ with $s_0$ the empty function and each $s_{k+1}$ extending $s_k$, so each $s_k$ is an injection $k\to A$ and $b_k:=s_{k+1}(k)$ defines an injection $\mathbb{N}\to A$ whose range is a countably infinite subset of $A$.

For claim 4 assume countable choice, let $X$ be limit point compact with every singleton closed, and let $\mathcal{U}$ be an at most countable open cover of $X$ with no finite subcover; as at step 1.4 the family is nonempty, [L7] lists it as $(U_n{)}_{n\in \mathbb{N}}$, the sets $E_n:=X\setminus (U_0\cup \cdots \cup U_n)$ are nonempty, and [L5] supplies a sequence $(x_n)$ with $x_n\in E_n$ for every $n$.

Sequential compactness gives a strictly increasing $j\mapsto n_j$ and $p\in X$ with $x_{n_j}\to p$; some $U_m$ contains $p$ and is a neighbourhood of it by [L4], so $x_{n_j}\in U_m$ for all large $j$, while $n_j\ge j$ by [L8] gives $n_j\ge m$ for all large $j$ and hence $x_{n_j}\mathrm{\notin }{U}_0\cup \cdots \cup U_{n_j}\supseteq U_m$ for those $j$ — impossible. So no such $\mathcal{U}$ exists and $X$ is countably compact, which is claim 2.

The sets $C_n$ satisfy ${\bigcap }_{n\in \mathbb{N}}{C}_n=\mathrm{\varnothing }$: a point outside $B$ lies in no $C_n$, and $b_k\mathrm{\notin }{C}_{k+1}$ because $k\mapsto b_k$ is injective. So $\mathcal{V}:=\{X\setminus C_n:n\in \mathbb{N}\}$ is an at most countable family of open sets whose union is $X$.

The set $A:=\{x_n:n\in \mathbb{N}\}$ of step 2.1 is infinite. Were it finite, then for each $a\in A$ the least $m$ with $a\in U_m$ exists, since $(U_n)$ covers $X$, and the largest $M$ of those finitely many least indices exists; but $x_M\in E_M$ misses $U_0\cup \cdots \cup U_M$ while $x_M\in A$ lies in $U_m$ for some $m\le M$.

Countable compactness applied to $\mathcal{V}$ gives a finite subfamily $V_0,\dots ,V_p$ with union $X$; each $V_j$ is $X\setminus C_m$ for some $m$, and taking $N_j$ to be the least such $m$ and $N$ the largest of $N_0,\dots ,N_p$ gives $V_j=X\setminus C_{N_j}\subseteq X\setminus C_N$ for every $j$, since the $C_n$ decrease. Hence $X=X\setminus C_N$ and $C_N=\mathrm{\varnothing }$, contradicting $b_N\in C_N$. So a countably infinite subset of a countably compact space has a limit point in it, which is claim 1(d).

Limit point compactness gives a limit point $p$ of the infinite set $A$; some $U_m$ contains $p$, the set $F:=\{x_0,\dots ,x_m\}\setminus \{p\}$ is closed by [L4] as a union of finitely many closed singletons, and $W:=U_m\setminus F$ is therefore open and contains $p$, hence is a neighbourhood of $p$ meeting $A\setminus \{p\}$: there is $n$ with $x_n\in W$ and $x_n\mathrm{\ne }p$.

Claim 3: given an infinite $A\subseteq X$ with $X$ countably compact, step 1.6 produces a countably infinite $B\subseteq A$, step 3.2 gives $B$ a limit point $p$ in $X$, and $p$ is then a limit point of $A$ by [L3], since $B\subseteq A$. So $X$ is limit point compact.

If $n\le m$ then $x_n$ is one of $x_0,\dots ,x_m$ and differs from $p$, so $x_n\in F$, contradicting $x_n\in W=U_m\setminus F$; hence $n>m$, so $U_m\subseteq U_0\cup \cdots \cup U_n$ and $x_n\in W\subseteq U_m$ contradicts $x_n\in E_n$. No such $\mathcal{U}$ exists, so $X$ is countably compact, which is claim 4.

Claims 1(a), 1(b), 1(c) and 1(d) are steps 1.1, 1.2, 1.3 and 3.2; claim 2 is step 2.2; claim 3 is step 4.2; and claim 4 is step 5.1.