Question about Theorem 2.24 in Baby Rudin 
Theorem 2.24.
(a) For any collection ${G_\alpha}$ of open sets, $\cup_\alpha G_\alpha%$ is open.
(b) For any coll ection ${F_\alpha}$ of closed sets, $\cap_\alpha F_\alpha%$ is closed.
(c) For any finite collection $G_1,...,G_n$ of open sets, $\cap_{i=1}^n G_i%$ is open.
(d) For any fi nite collection $F_1,...,F_n$  of closed sets, $\cup_{i=1}^n F_i%$ is closed.

Here is the proof of (c) in the book.

Proof.$\quad$  Put  $H=\cap_{i=1}^n G$. For any $x\in H$,  there exist neighborhoods
$N_i$ of $x$, with radii $r_i$  such  that  $N\subset G_i(i=1,...,n)$. Put
$r=\min(r_1,...,r_n)$, and let $N$ be the neighborhood of $x$ of radius $r$. Then  $N\subset G_i$ for $i=1,...,n$, so that $N\subset H$,  and $B$ is open.

So, why can't we use the similar process to prove $\cap_{i=1}^\infty G_i%$ is open? Can't we get the $r=\min(r_1,...,r_n,...)$ when $n$ is for all the natural numbers?
 A: Actually the "similar process" mentioned by you is not similar.
For a collection of arbitrary sets $G$s, you can only know that $\min$ always exists if there are finitely many (because you could check each one of them, and find the min one) - however you cannot do so if there are infinitely many. Thus you do not know if $\min$ exists or not (but we could say about its infimum, which does not help because we want to find a G out of the collection).
A: *

*mind that there is a difference between the $infimum$ and the
$minimum$ of a set. For instance, take $$A:=\{\frac{1}n; n \in
\mathbb{N} \} \subset \mathbb{R}$$ This set does not have a minimal
element, since $0 \notin A$, but $inf(A)=0$. If the $min$ exists, it
is the same as the $inf$, but the existance of the latter does not
guarantee the existence of the first. So in your set of indices
$\{r_1,r_2,\dots\}$ there is not necesserily a minimal element.

*Another (counter-) example why your proposition is not true: Consider $$B:=\bigcap_{k=1}^{\infty}
B_k(0) \subset \mathbb{R}^n,$$ where $B_k(0):=\{x \in \mathbb{R}^n:|x|< 1+1/k, \space k \in
\mathbb{N}\} $. Cleary, $B_k(0) $ is open for every $k \in
\mathbb{N}$, but $B= \{x \in \mathbb{R}^n:|x|\leq 1 \}$ is the closed
unit ball and therefore not open.
A: Well, a trivial but critical requirement is that $r$ exists. However, an arbitrary subset of real numbers doesn't necessarily have a minimum, unless provided that it's finite. That is to say, $\min(r_1,r_2,r_n,...)$ doesn't necessarily exist. So the proof in case of $G$'s finite couldn't be extended to the cases when $G$'s are infinite.
Actually, extending a theorem is rather risky, unless you clearly understand the theorems used in the proof.
A: You can be looking for a min forever, so it's very different to have finitely many subsets or infinite subsets. You need to prove that this minium exists before apply this way to prove it.
This will help you to understand why we cannot assume the existence of such min: 
Maximum Number in an Infinite set
