Clarification on " the collection of subsequential limits of a sequence in a metric space form a closed subset" 
This theorem is from Rudin. My question is can we avoid that $\delta$? instead of it, can we just take $2^{-i}$? I think that would still work. If we can't avoid that $\delta$, then please justify.
[I'm reading Rudin on my own and I'm a beginner, so don't mind my trivial question]

Yes, you can use $2^{-i}$, though it’s slightly more elegant to use $2^{-(i+1)}$. At stage $i$ of the construction you can argue that since $q$ is a limit point of $E^*$, there is an $x\in E^*$ such that $d(x,q)<2^{-(i+1)}$, and since $x\in E^*$, there is an $n_i>n_{i-1}$ such that $d(x,p_{n_i})<2^{-(i+1)}$. It follows that $$d(q,p_{n_i})<2^{-(i+1)}+2^{-(i+1)}=2^{-i}$$ for $i\ge 2$ and hence that $\langle p_{n_i}:i\ge 1\rangle$ converges to $q$, so that $q\in E^*$.
