In any metric space compactness implies countable compactness and limit point compactness, and each of countable compactness and limit point compactness implies sequential compactness; every implication here is proved without a choice principle

Claim 1 is immediate: an at most countable family of open sets with union $X$ is in particular a family of open sets with union $X$, so compactness supplies the finite subfamily that countable compactness asks for.

For claim 2, assume $(X,d)$ compact, let $A\subseteq X$ have no limit point in $X$, and put $\mathcal{U}:=\{U\subseteq X:U\text{ open in }X\text{ and }U\cap A\text{ has at most one element}\}$, a family cut out by a property; $\mathcal{U}$ has union $X$, because each $p\in X$ fails to be a limit point of $A$ and so admits $r>0$ with $B(p,r)\cap (A\setminus \{p\})=\mathrm{\varnothing }$, whence $B(p,r)\in \mathcal{U}$ and $p\in B(p,r)$.

Compactness gives $n\in \mathbb{N}$ and $U_0,\dots ,U_n\in \mathcal{U}$ with $X=U_0\cup \cdots \cup U_n$, unless $X=\mathrm{\varnothing }$, in which case $A=\mathrm{\varnothing }$ is finite.

Define $\phi :A\to \sigma (n)$ by letting $\phi (a)$ be the least $i\le n$ with $a\in U_i$; this is well defined and canonical, and it is injective, since $\phi (a)=\phi (b)=i$ puts both $a$ and $b$ in $U_i\cap A$, a set with at most one element. Hence $A$ is in bijection with $\phi [A]$, a subset of $\mathbb{N}$ bounded above by $n$, so $A$ is finite.

So a subset of $X$ with no limit point in $X$ is finite; contrapositively every infinite subset of $X$ has a limit point in $X$, and $(X,d)$ is limit point compact: claim 2.

For claim 3, assume $(X,d)$ countably compact, let $(x_k)$ be a sequence in $X$ and put $T_n:=\overline{\{x_k:k\ge n\}}$ for $n\in \mathbb{N}$.

Each $T_n$ is closed and contains $x_n$, and $T_m\subseteq T_n$ whenever $m\ge n$, because $\{x_k:k\ge m\}\subseteq \{x_k:k\ge n\}\subseteq T_n$ and $T_m$ is the smallest closed superset of the first set.

Suppose for contradiction that ${\bigcap }_{n\in \mathbb{N}}{T}_n=\mathrm{\varnothing }$; then $\mathcal{V}:=\{X\setminus T_n:n\in \mathbb{N}\}$ is an at most countable family of open subsets of $X$ whose union is $X\setminus {\bigcap }_n{T}_n=X$.

Countable compactness gives a finite subfamily $V_0,\dots ,V_p$ of $\mathcal{V}$ with union $X$; putting $W_i:=X\setminus V_i$, each $W_i$ equals $T_n$ for at least one $n$, so finite choice applied to $i\mapsto \{n\in \mathbb{N}:T_n=W_i\}$ yields indices $n_0,\dots ,n_p$ with $W_i=T_{n_i}$, and a greatest member $n_j$ of that list satisfies $T_{n_j}\subseteq T_{n_i}$ for every $i\le p$.

Then $T_{n_j}=T_{n_0}\cap \cdots \cap T_{n_p}=X\setminus (V_0\cup \cdots \cup V_p)=\mathrm{\varnothing }$, contradicting $x_{n_j}\in T_{n_j}$.

Hence there is $p\in X$ with $p\in T_n$ for every $n\in \mathbb{N}$.

For every $k\in \mathbb{N}$ and every $m\in \mathbb{N}$ the set $\{j\in \mathbb{N}:j>m\text{ and }d(x_j,p)<1\mathrm{/}(k+2)\}$ is nonempty, since $p\in T_{m+1}$ means that the ball $B(p,1\mathrm{/}(k+2))$ meets $\{x_j:j\ge m+1\}$; so it has a least element, and likewise $\{j:d(x_j,p)<1\}$ is nonempty and has a least element $m_0$.

Applying recursion on $\mathbb{N}\times \mathbb{N}$ to the starting value $(0,m_0)$ and the rule $f(k,m):=(k+1,\text{ the least }j>m\text{ with }d(x_j,p)<1\mathrm{/}(k+2))$ produces $g:\mathbb{N}\to \mathbb{N}\times \mathbb{N}$ whose first coordinate at $k$ is $k$; write $n_k$ for its second coordinate.

Then $n_k<n_{k+1}$ for every $k$, so $k\mapsto n_k$ is strictly increasing, and $d(x_{n_k},p)<1\mathrm{/}(k+1)$ for every $k$, the case $k=0$ being the choice of $m_0$.

Given a rational $\epsilon >0$ take a natural $N\ge 1$ with $1\mathrm{/}N<\epsilon$; for $k\ge N$ one has $k+1>N$ and so $d(x_{n_k},p)<1\mathrm{/}(k+1)<1\mathrm{/}N<\epsilon$. Hence $x_{n_k}\to p$, the sequence $(x_k)$ has a convergent subsequence, and $(X,d)$ is sequentially compact: claim 3.

For claim 4, assume $(X,d)$ limit point compact, let $(x_k)$ be a sequence in $X$ and let $R:=\{x_k:k\in \mathbb{N}\}$ be its range, a nonempty subset of $X$.

Suppose first that $R$ is finite, and list it as $R=\{v_0,\dots ,v_m\}$; putting $S_i:=\{k\in \mathbb{N}:x_k=v_i\}$ for $i\le m$ gives $\mathbb{N}=S_0\cup \cdots \cup S_m$.

Some $S_i$ is unbounded in $\mathbb{N}$: otherwise each $S_i$ has an upper bound in $\mathbb{N}$ and hence a least upper bound $N_i$, canonical by well-ordering, and a greatest member $N$ of the list $N_0,\dots ,N_m$ would satisfy $N+1>N_i$ for every $i$, so that $N+1$ lies in no $S_i$, against $\mathbb{N}=S_0\cup \cdots \cup S_m$. Let $i^{\ast }$ be the least $i\le m$ for which $S_i$ is unbounded.

Recursion applied to the starting value $\min S_{i^{\ast }}$ and the rule $f(m):=\min \{k\in S_{i^{\ast }}:k>m\}$, each of these sets being nonempty because $S_{i^{\ast }}$ is unbounded, produces a strictly increasing $k\mapsto n_k$ with $x_{n_k}=v_{i^{\ast }}$ for every $k$; a constant sequence converges to its value, so $x_{n_k}\to v_{i^{\ast }}\in X$.

Suppose instead that $R$ is infinite; limit point compactness then gives a limit point $p\in X$ of $R$.

Suppose for contradiction that some real $r>0$ and some $N\in \mathbb{N}$ satisfy $d(x_k,p)\ge r$ for every $k\ge N$.

Let $E$ be the set listed by $r$ together with the $N$ entries $e_k$ for $k<N$, where $e_k:=d(x_k,p)$ if $x_k\mathrm{\ne }p$ and $e_k:=r$ otherwise; every listed entry is a positive real, so $E$ is a nonempty finite set of positive reals and $s:=\min E>0$. Then $B(p,s)$ misses $R\setminus \{p\}$: a point of $R\setminus \{p\}$ is $x_k$ with $x_k\mathrm{\ne }p$, and $d(x_k,p)\ge r\ge s$ when $k\ge N$, while $d(x_k,p)=e_k\ge s$ when $k<N$. That contradicts $p$ being a limit point of $R$.

Hence for every real $r>0$ and every $N\in \mathbb{N}$ there is $k\ge N$ with $d(x_k,p)<r$.

Consequently, for every $k$ and every $m$ the set $\{j>m:d(x_j,p)<1\mathrm{/}(k+2)\}$ is nonempty, as is $\{j:d(x_j,p)<1\}$, and the recursion of steps 13.1 and 14.1 applies verbatim, producing a strictly increasing $k\mapsto n_k$ with $d(x_{n_k},p)<1\mathrm{/}(k+1)$; by the estimate of step 15.1, $x_{n_k}\to p$.

In both cases $(x_k)$ has a subsequence converging in $X$, so $(X,d)$ is sequentially compact: claim 4.

Claims 1, 2, 3 and 4 are proved by steps 1.1, 5.1, 15.1 and 25.1 respectively.