My question is about exactly how small can a nbhd of $x^\ast$ be, or put in another way: is it possible that some subcover $G_\alpha$ be constructed along the way so that it each time (albeit narrowly) "escapes"? Rudin says "Since $G_{\alpha}$ is open , there exist $r >0$ such that $|y-x^\ast|<r $ implies that $y \in G_{\alpha}$", fair enough, but why assume then that such an $r$ is fixed so that some very large $n$ can in some way exceed it? After all some $r^\prime \in (0,2^{-n}\delta)$ can always be found s.t. $ Nr(x^\ast) \subset G_\alpha$ iff $r \leq r^\prime$, provided that we are not required to describe $G_\alpha$ beforehand.
Theorem 2.40 : Every k-cell is compact.
Proof: Let $I$ be a $k$-cell consisting of points $\bf x$ with $a_j\le x_j\le > b_j$. Put
$$ \delta =\left (\sum_1^k(b_j-a_j)^2\right )^{1/2}$$
Then $|x-y|\le \delta, \forall x,y \in I$.
Suppose, to get a contradiction, that there exists an open cover $\{G_{\alpha}\}$ which has no finite subcover of $I$.
Put $c_{j}=(a_{j}+b_{j})/2$. The intervals $[a_{j},c_{j}]$ and $[c_{j},b_{j}]$ then determine $2^{k}$ $k$-cells $Q_{i}$ whose union is $I$.
One of these $Q_i$, call it $I_1$, will also have no finite subcover. Continuing this process we obtain a sequence $I_n$ with the following properties:
$I \supset I_1 \supset I_2 \supset I_3 \supset I_4 \supset \dots$.
$I_n$ is not covered by any finite subcollection of $\{ G_{\alpha}\}$.
If $x,y \in I_n$, the $|x-y| \leq 2^{-n}\delta$
by $( a)$ and theorem $2.39$ , there is a point $x^\ast$ which lie in every $I_n$. For some $\alpha , x^\ast \in G_\alpha.$ Since $G_{\alpha}$ is open , there exist $r >0$ such that $|y-x^\ast|<r $ implies that $y \in G_{\alpha}$. If $n$ is so large that $2^{-n} \delta < r$ (there is such an $n$ , for otherwise $2^n \le \delta/r$ for all positive integr $n$ , which is absurd since $R$ is archimedean , then $(c)$ implies that ${I_n \subset G_{\alpha}}$ , which contradicts $(b)$.
A similar argument is used in theorem 2.44 (Cantor set), in a more explicit form by this answer:
Theorem: The Cantor set $P$ contains no segment $(\alpha,\beta)$ with $\alpha < \beta$.
Proof: By definition, $E_n$ does not contain any segments of length greater than $3^{-n}$. For some $m > 0$, the segment $(\alpha,\beta)$ satisfies $\beta - \alpha > 3^{-m}$ and so $(\alpha,\beta) \not\subset E_m$ and therefore, $(\alpha,\beta) \not\subset \bigcap_{n=1}^{\infty} E_n = P$. $_\Box$