I want to present my alternative proof to theorem 3.7 from Baby Rudin.
Either it is correct and thus Rudins proof is just a bit more complicated but equal in power, or my proof is wrong and I am missing some detail from Rudins proof.
So theorem states:
The subsequential limits of a sequence $ \{p_n\} $ in a metric space $ X $ form a closed subset of $ X $
My strategy is exactly the same as Rudins:
Take limit point $ q $ of $ E^* $, where $ E^* $ is a set of said subsequential limits, and show that $ q $ is a limit of some subsequence of $ \{p_n\} $. The difference will be in how the subsequence is constructed.
Since $ q $ is a limit point, there exists $ x \in E^* $ such that $ d(q, x) < 1 $, where $ 1 $ was chosen arbitrarily.
But then $ 0 < 1 - d(q, x) $, so, since $ x $ is a limit of some subsequence of $ \{p_n\} $, there exists $ n_1 $ such that $$ d(x, p_{n_1}) < 1 - d(q, x) $$ so: $$ d(q, p_{n_1}) \leq d(q, x) + (x,p_{n_1}) < 1 $$
Now assume we have already chosen $ p_1, ..., p_{n_{k-1}} $.
We find $ x \in E^* $ such that $ d(q, x) < \frac{1}{k} $.
We can find $ n_k $ such that $$ d(x, p_{n_k}) < \frac{1}{k} - d(q, x) $$
so, as previously, we get:
$$ d(q, p_{n_k}) < \frac{1}{k} $$
Since there is infinitely many such $ p_k $ (there is infinitely many of them around limit $ x $), we can get $ n_{k-1} < n_k $.
We thus obtain subsequence $ p_{n_1}, p_{n_2}, ... $, which obviously converges to $ q $, since we can make $ \frac{1}{k} $ as small as we want.
My questions
Is the basic idea of proof correct?
Are there any important details I have missed, which I should clarify?
Is there a reason why Rudin decides to divide distance by two between every iteration of choosing next $ p_{n_k} $, or is it simply another way to get a sequence that tends to $ 0 $ as $ n $ get larger?
Yes, your basic idea is correct.
No, there are not important details that you have missed or should clarify.
Not really. As you have probably understood, one may choose any sequence $a_k$ tending to zero. You chose $1/k$; Rudin chose $2^{-k}$. Between these two choices, there is no special advantage in choosing one or the other. The choice $2^{-k}$ is a kind of universal choice because it can also be summed up and converge, if necessary. In this proof it is not necessary, but somewhat customary.