Def'n 1. is the general def'n of compactness in topology, whether or not the topology can be generated by a metric.
In a metric space, a set $X$ is compact iff every sequence in $X$ has a limit point that BELONGS to $X$ iff every infinite subset of $X$ has a limit point that belongs to $X.$
The n-dimensional generalization of the (1-dimensional) Bolzano-Weierstrass theorem is that a subset of $R^n$ is compact iff it is closed and bounded. (But this does not hold for all metric spaces.)
It is useful to know various equivalents to compactness, and to know that there are equivalents specific to metric spaces, just as it is useful to know various equivalents of "continuous function", some of which are specific to metric spaces.
Another term is "pre-compact" which I have seen only in the context of Hausdorff spaces : $X$ is pre-compact iff $\overline X$ is compact.
Lemma: Let $(X,d)$ be a metric space. For each $n \in \{1,2,3,\ldots\}$ there is (by Zorn's lemma) a maximal (by inclusion) set $D_n$ such that for all $x,y \in D_n$ with $x \neq y$, we have $d(x,y) \ge \frac{1}{n}$. If all $D_n$ are at most countable, then $X$ is separable.
Proof: We'll show that $D = \cup_n D_n$ is dense (and is countable when all $D_n$ are), and so pick any $x \in X$ and any $r>0$ and we'll show that $B(x,r) \cap D \neq \emptyset$. Suppose not, then find $m$ with $\frac{1}{m} < r$. Then from $B(x,r) \cap D = \emptyset$ we know that
$d(x, y) \ge \frac{1}{m}$ for all $y \in D_m$ (or else $y \in D \cap B(x,r)$), and then $D_m \cup \{x\}$ would contradict the maximality of $D_m$. So the intersection with $D_m$ and hence $D$ is non-empty, and so $D$ is dense.
(The above is the heart of the proof that a ccc metric space is separable, as $X$ ccc implies that all $D_m$ are countable, as the $B(x,\frac{1}{m})$, $x \in D_m$ are a pairwise disjoint open family of non-empty sets.)
Now suppose that $(X,d)$ is countably compact. Suppose for a contradiction that $X$ is not separable. Then for some $m \ge 1$ we have that $D_m$ (as in the lemma) is uncountable.
Fix such an $m$.
Now $D_m$ is discrete (clear as $B(x,\frac{1}{m}) \cap D_m = \{x\}$ for each $x \in D_m$) and closed: suppose that $y \in X\setminus D_m$ is in $\overline{D_m}$, then $B(y, \frac{1}{2m})$ contains infinitely many points of $D_m$ and for any $2$ of them, say $x_1, x_2 \in D_m$, we'd have $d(x_1, x_2) \le d(x_1, y) + d(y,x_2) < \frac{1}{m}$ contradiction (as points in $D_m$ are at least $\frac{1}{m}$ apart). So $D_m$ is closed and discrete (as an aside: being uncountable this would already contradict Lindelöfness of $X$; this shows that a Lindelöf metric space is separable, e.g.), but we want to contradict countable compactness, which is easy too:
Choose $A \subseteq D_m$ countably infinite (so that $A$ is closed in $X$, as all subsets of $D_m$ are), and define a countable open cover $\mathcal{U} = \{B(p, \frac{1}{m}): p \in A\} \cup \{X\setminus A\}$ of $X$ that has no finite subcover: we need every $B(p,\frac{1}{m})$ to cover $p$ for all $p \in A$.
This contradiction then shows $X$ is separable (and thus has a countable base ,is Lindelöf etc. finishing the compactness).
The crucial fact is that all of the following are equivalent for a metric space:
- $X$ has a countable base.
- $X$ is separable.
- All discrete subspaces of $X$ are at most countable.
- All closed and discrete subspaces of $X$ are at most countable.
- $X$ is ccc.
- $X$ is Lindelöf.
In the above I essentially did "not (2) implies not (4)" implicitly. The fun is that countable compactness implies (4) easily (such closed discrete subspaces are even finite) and thus we get Lindelöfness "for free". Also implicit in the above proof is that every countably compact space is limit point compact.
Best Answer
I already showed in this answer that $X'$ relatively compact implies $X'$ relatively countably compact. Also, if $X'$ is relatively compact, the $\overline{X'}$ is compact, and thus sequentially compact (this holds in particular in metric spaces, but also more broadly). So any sequence from $X'$ has a convergent subsequence with limit in $\overline{X'}$, so $X'$ is then relatively sequentially compact as well.
In any space, $X'$ relatively sequentially compact implies $X'$ relatively countably compact: any infinite subset $A$ of $X'$ contains some sequence with all different elements, which has a convergent subsequence to some $x \in X$, and this $x$ is an accumulation point of $A$.
If $X'$ is relatively countably compact, and $X$ is metric (first countable and $T_1$ will already do), let $(x_n)$ be a sequence from $X'$. If $A = \{x_n: n \in \mathbb{N}\}$ is finite, some value occurs an infinite number of times, and yields a convergent subsequence. So assume $A$ is infinite, so it has an accumulation point $p \in X$. Because $X$ is $T_1$, this means that every neighbourhood of $p$ intersects $A$ in infinitely many points. Pick $x_{n_1}$ in $B(p, 1)$, $x_{n_2}$ with $n_2 > n_1$ in $B(p, \frac{1}{2})$, and so on, by recursion. This defines a convergent subsequence of $(x_n)$ that converges to $p$. So $X'$ is relatively countably compact.
If $X'$ is relatively countably compact, and $X$ is metric, then we do get that $X'$ is relatively compact (this is due to Hausdorff, IIRC). I cannot reconstruct a proof right away, but there is one here, e.g. (but this involves Cauchy filters etc.)
It turns out that these notions are also equivalent in some topological vector spaces (spaces of the form $C_p(X)$ where $X$ is compact (Grothendieck), and weak topologicals on normed vector spaces (Eberlein-Smulian)), but these are a bit more involved, I think.