What's going on with "compact implies sequentially compact"?

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.

Compactness is equivalent to sequential compactness for metric spaces; in general, neither implies the other. I believe that the point at which your proof requires that we be working in a metric space, is in

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 $(x_n)$

In a metric space, we can get this subsequence by taking balls of radius $\frac{1}{n}$; in general we can't.

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.