I believe that you are looking for ideas from the Cantor Bendixson theorem.
The main idea of the proof is the Cantor-Bendixson derivative. Given a closed set $X$, the derived set $X'$ consists of all limit points of $X$. That is, one simply throws out the isolated points. Continuing in a transfinite sequence, one constructs $X_\alpha$ as follows:
- $X_0=X$, the original set.
- $X_{\alpha+1}=(X_\alpha)'$, the set of limit points of $X_\alpha$.
- $X_\lambda=\bigcap_{\alpha\lt\lambda}X_\alpha$, for limit ordinals $\lambda$.
Thus, $X_1$ consists of the limit points of $X$, and $X_2$ consists of the limits-of-limits, and so on. The set $X_\omega$ consists of points that are $n$-fold limits for any particular finite $n$, and $X_{\omega+1}$ consists of limits of those kind of points, and so on. The process continues transfinitely until a set is reached which has no isolated points; that is, until a perfect set is reached. The Cantor Bendixon rank of a set is the smallest ordinal $\alpha$ such that $X_\alpha$ is perfect.
The concept is quite interesting historically, since Cantor had undertaken this derivative before he developed his set theory and the ordinal concept. Arguably, it is this derivative concept that led Cantor to his transfinite ordinal concept.
It is easy to see that the ordinal $\omega^\alpha+1$ under the order topology has rank $\alpha+1$, and one can use this to prove a version of your desired theorem.
The crucial ingredients you need are the Cantor Bendixson rank of your space and the number of elements in the last nonempty derived set. From this, you can constuct the ordinal $(\omega^\alpha+1)\cdot n$ to which your space is homeomorphic.
Meanwhile, every countable ordinal is homeomorphic to a subspace of $\mathbb{Q}$, and is metrizable. The compact ordinals are precisely the successor ordinals (plus 0).
Update 5/11/2011. This brief article by Cedric Milliet contains a proof of the Mazurkiewicz-Sierpiński theorem (see
Stefan Mazurkiewicz and Wacław Sierpiński, Contribution à la topologie des
ensembles dénombrables, Fundamenta Mathematicae 1, 17–27, 1920), as follows:
Theorem 4. Every countable compact
Hausdorff space is homeomorphic to some well-ordered set with the order topology.
The article proves more generally that any two countable locally compact Hausdorff spaces $X$ and $Y$ of same Cantor-Bendixson rank and degree are homeomorphic. This is proved by transfinite induction on the rank, and the proof is given on page 4 of the linked article.
A compact Hausdorff space need not be sequentially compact, there is an example here (near the end of the page), and a sequentially compact space need not be compact (e.g. the first uncountable ordinal $\omega_1$ in the order topology, which is first countable, hereditarily normal, sequentially compact, countably compact, but not compact).
Of course a compact space is trivially countably compact: if all open covers have finite subcovers, all countable ones certainly do.
A sequence $(x_n)$ (say of all distinct terms) in a countably compact space $X$ can be seen as an infinite set, and so it has an $\omega$-accumulation point (every neighbourhood of $p$ contains infinitely many terms of the sequence). See my answer here.
But this does not guarantee that there is a subsequence that converges to $p$ (see the proof in the link I gave), because there can be a lot of neighbourhoods of $p$, and we need every neighbourhood of $p$ to contain a tail of the subsequence.
If however $p$ has a countable local base $O_n$ (like in metric spaces, where we can use the balls around of $p$ of radius $\frac{1}{n}$), then we can find such a subsequence: pick $n_1$ with $x_{n_1} \in O_1$, $n_2 > n_1$ with $x_{n_2} \in O_1 \cap O_2$, $n_3 > n_2$ such that $x_{n_3} \in O_1 \cap O_2 \cap O_3$ and so on. This uses that all these (finite!) intersections are neighbourhoods of $p$ and thus contain infinitely many terms of the sequence. And the fact that these $O_n$ form a local base, ensures that the subsequence $x_{n_k}$ converges to $p$ as $k$ tends to infinity: if $O$ is open and contains $p$, for some $m$, $O_n \subset O$, and so all $x_{n_k}$ with $k \ge m$ are by construction in $O_m$ and hence in $O$.
So in first countable (countably) compact spaces we do have sequential compactness, and
many spaces are indeed first countable (like metrisable spaces), so most people's intuition is more tuned to that case. But in general we do not have that compact spaces are sequentially compact.
Best Answer
The paper by J. Novák in my second link at your second link uses an older terminology: what he calls compact is our countably compact, and for our compact he uses the term bicompact. The example in his Theorem $4$ is an example of two countably compact Tikhonov spaces whose product is not countably compact. It is also an example of two Tikhonov limit point compact spaces whose product is not limit point compact.