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.
There are limit point compact spaces that are not compact, even uniformisable spaces, e.g. take $X=\omega_1$ (the first uncountable ordinal in the order topology (that Munkres calls $W$ IIRC)). $X$ is also sequentially compact.
I don’t see how adding an equivalent formulation (filter compactness), which is equivalent in all topological spaces to standard (open cover ) compactness, is going to change anything. $X$ is not compact so it’s not filter compact, period. If $S$ is any uniform space, being limit point compact as a topological space is not going to make it compact.
Both the equivalence of compact and filter compact for all topological spaces and the equivalence of compactness with “complete and totally bounded” in all uniform spaces are classic and well-known. What the OP states is just a combination of these two facts.
Sequential compactness “orthogonal” to compactness: one need not imply the other in general spaces. “Filter compactness” is just another way to say compactness, don’t let an analogy with sequences mislead you here. “Net compactness “ (every net has a convergent subnet”) is also just a reformulation of compactness for all spaces. Sequences are “special”.
Best Answer
One of my favorite textbooks is Klaus Janich's Topology, and he has a nice motivation for compactness I feel, namely why we should care about. This is in addition to my comment about compact subsets of a Hausdorff space being essentially like finite point sets. But he writes:
This is nice, but it is slightly advanced, and he gives some examples that follow like a continuous/locally bounded map from a compact space to $\mathbb{R}$ is bounded, and some discussions of locally finite covers and manifolds (honestly, I like this book after the fact of learning topology, not to learn from).
Hope that helps somewhat.