TITLE: What's going on with "compact implies sequentially compact"?
QUESTION [79 upvotes]: I've seen both counterexamples and proofs to "compact implies sequentially compact", and I'm not sure what's going on.
Apparently there are compact spaces which are not sequentially compact; quick googling and wikipedia checks will turn up examples floating around; they tend to be variants of $[0,1]^{[0,1]}$ with the product topology.  Here's a demonstration(?):

$[0,1]^{[0,1]}$ is compact by Tychonoff's theorem.  So, we demonstrate failure of sequential compactness.
Choose a unique binary representation for each $x\in [0,1]$.
For each $n\in\mathbb{N}$, let $f_n : [0,1]\to[0,1]$ (an element of $[0,1]^{[0,1]}$) be the function which maps each $x$ to its $n$-th place digit in its binary expansion.
Let $f_{n_k}$ be a subsequence of this sequence.
Let $x'\in [0,1]$ be such that the $n_{2m}$-th digit is $0$ and the $n_{2m+1}$-th digit is $1$, for all $m\in\mathbb{N}$.
Then $f_{n_k} (x')$ does not converge (it alternates between $0$ and $1$), and hence $f_{n_k}$ cannot converge.
Thus we have found a sequence in $[0,1]^{[0,1]}$ without any convergent subsequence and so it is not sequentially compact.

(Aside: this is apparently based off of an example in Steen's Counterexamples in Topology, according to http://ncatlab.org/nlab/show/sequentially+compact+space )
Nevertheless, there's also some proofs(?) floating around that compactness of a space implies sequential compactness, along these lines (this proof(?) is of the contrapositive):

Suppose $X$ is not sequentially compact.
By definition, this means there is some sequence $(x_n)$ over $X$ with no convergent subsequence.
If any $x\in X$ had for each of its neighborhood $U$ infinitely many $n$ for which $x_n \in U$, then we could define a convergent subsequence of $(x_n)$, contradicting our assumption.  (Presumably this is done by choosing for each neighborhood a sufficiently-large-indexed term in that neighborhood.)
Thus, for each $x \in X$ we can select an open set $U_x$ such that $x\in U_x$ but with $x_n \in U_x$ for only finitely many $n$.
The collection $\mathcal{U}=\{U_x : x\in X\}$ is clearly an open cover of $X$.
If $\mathcal{U}$ had a finite subcover $\{U_1, \dots, U_k\}$ then the union $U_1 \cup \cdots \cup U_k$ would contain all of $X$ but only contain $x_n$ for finitely many $n$, which is impossible.
Thus $X$ is not compact, since we have found an open cover without a finite subcover.

(Aside: this proof is essentially the same as the proof appearing in Rudin's Principles of Mathematical Analysis for Thm 2.37 that infinite subsets of compact spaces have limit points.)
So, what's going on here?  It can't be that what both the counterexample and the proof are telling us is correct.  Is there some subtle (or more embarassingly for me -- glaring) flaw in the proof?

REPLY [17 votes]: In general spaces we have the problem, as pointed out by others, that the topology need not be "determined by sequences", like we have in metric spaces, or first countable spaces. A class of spaces where we do have that sequences suffice is the class of sequential spaces, which turn out to be the class of quotient spaces of metric spaces. Here we have that sequential closure always equals closure, and for Hausdorff sequential spaces we can prove
(e.g. see this blog post) quite easily that sequential compactness is equivalent to countable compactness. The latter is always implied by compactness, so that for sequential spaces we have compact implies sequentially compact (but not reversely, as $\omega_1$ in the order topology shows).
Another classical example of a compact Hausdorff but not sequentially compact space is the Čech-Stone compactification of the integers. Such spaces fail sequential compactness because they are too big, as there is a so-called cardinal invariant $\omega_1 \le t \le c$ such that every compact Hausdorff space $X$ of size $\lt 2^t$ is always sequentially compact.