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?

The problem is here:


  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.)


In general topological spaces this only implies that we are able to construct a convergent net, not a convergent sequence. (A point $x$ is an accumulation point of a subset $S$ $\Leftrightarrow$ there exists a net of points of $S\setminus\{x\}$ converging to $x$.)

If $X$ is first countable at $x$ (the point $x$ has a countable base), then a sequence can be constructed. (This is more-or-less standard. We first construct a decreasing base $U_n$ at $x$ and then choose a point from each $U_n$). In particular, this works for metric spaces. Note that Rudin works only with compact subset of metric spaces in that chapter.
