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 $B(p,r) \cap (A \setminus \{p\}) = \emptyset$, 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 $X = U_0 \cup \dots \cup U_n$, unless $X = \emptyset$, in which case $A = \emptyset$ is finite.

Define $\varphi : A \to \sigma(n)$ by letting $\varphi(a)$ be the least $i \le n$ with $a \in U_i$; this is well defined and canonical, and it is injective, since $\varphi(a) = \varphi(b) = i$ puts both $a$ and $b$ in $U_i \cap A$, a set with at most one element. Hence $A$ is in bijection with $\varphi[A]$, 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 $\bigcap_{n \in \mathbb{N}} T_n = \emptyset$; 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 $T_{n_j} = T_{n_0} \cap \dots \cap T_{n_p} = X \setminus (V_0 \cup \dots \cup V_p) = \emptyset$, 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 $\{\, j \in \mathbb{N} : j > m \text{ and } d(x_j,p) < 1/(k+2) \,\}$ is nonempty, since $p \in T_{m+1}$ means that the ball $B(p, 1/(k+2))$ 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 $f(k,m) := (k+1,\ \text{the least } j > m \text{ with } d(x_j,p) < 1/(k+2))$ 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 $d(x_{n_k},p) < 1/(k+1)$ for every $k$, the case $k = 0$ being the choice of $m_0$.

Given a rational $\varepsilon > 0$ take a natural $N \ge 1$ with $1/N < \varepsilon$; for $k \ge N$ one has $k + 1 > N$ and so $d(x_{n_k},p) < 1/(k+1) < 1/N < \varepsilon$. 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 $\mathbb{N} = S_0 \cup \dots \cup S_m$.

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 $\mathbb{N} = S_0 \cup \dots \cup S_m$. 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 $x_k \ne p$ 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 $x_k \ne p$, 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 $\{\, j > m : d(x_j,p) < 1/(k+2) \,\}$ 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 $d(x_{n_k},p) < 1/(k+1)$; 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.