Why the $\delta$ in this proof (Rudin 3.7)? I am confused on why it is necessary to define the sequence converging to $q$ in this manner in Rudin's theorem 3.7?:

Proof $\ \ $ Let $E^*$ be the set of all subsequential limits of $\{p_n\}$ and let $q$ be a limit point of $E^*$. We have to show that $q\in E^*$.
$\qquad$ Choose $n_1$ so that $p_{n_1}\neq q$. (If no such $n_1$ exists, then $E^*$ has only one point, and there is nothing to prove.) Put $\delta=d(q,p_{n_1})$. Suppose $n_1,...,n_{i-1}$ are chosen. Since $q$ is a limit point of $E^*$, there is an $x\in E^*$ with $d(x,q)<2^{-i}\delta$. Since $x\in E^*$, there is an $n_i>n_{i-1}$ such that $d(x,p_{n_i})<2^{-i}\delta$. Thus $$d(q,p_{n_i})\leq 2^{1-i}\delta$$ for $i=1,2,3,...$. This says that $\{p_{n_i}\}$ converges to $q$. Hence $q\in E^*$.

For instance, why not just start from the $5$th line and say:Since $q$ is a limit point of $E^*$, for each $i \in \mathbb{N}$, there is an $x \in E^*$ such that $d(x,q)<2^{-i}$. Since $x \in E^*$, there is an $n_i>n_{i-1}$ such that $d(x,p_{n_i})<2^{-i}$. So that
$$d(p_{n_i},q)<2^{1-i}$$
So that $\{p_{n_i}\}\rightarrow q$?
What is the point of setting $\delta=d(q,p_{n_1})$?Why even have it as a factor in $2^{1-i}\delta$?Is it necessary for the induction?
 A: 
for each $i∈N$, there is an $x∈E^∗$ such that $d(x,q)<2^{−i}$. Since $x∈E^∗$, there is an $n_i>n_i−1$ such that $d(x,p_{n_i})<2^{-i}$. So that
$d(p_{n_i},q)<2^{-i}$
So that ${p_{n_i}}→q?$

What if your chosen $x$ turns out to be $q$? Then your argument becomes circular.
A: Here is an alternative proof of that fact without that admittedly weird $\delta.$
Theorem. The set of limit points of a sequence $(p_n)$ is closed.
Proof. Let $E'$ be the set of limit points of the sequence $(p_n)$ (i.e. the set of all subsequential limits). Suppose $q$ is a limit point of $E'.$ By definition, there is a sequence $(q_i)$ defined in $E'$ such that $d(q_i, q) \to 0.$ Thus, by choosing a subsequence should the need arise, we may assume that $d(q_i, q) < \dfrac{1}{2^{i+1}}.$ Now, $q_i \in E',$ so by definition, there is a strictly increasing $\varphi_i:\mathbf{N} \to \mathbf{N}$ such that $p_{\varphi_i(n)} \to q_i.$ Define now $N_1$ to be the first natural number such that all $n \geq N_1$ satisfy $d(p_{\varphi_1(n)}, q_1) < \dfrac{1}{2^{1+1}}$ (such $N_1$ exists by the definition of convergence). Suppose that $N_1 < \ldots < N_{i-1}$ have been defined. Define $N_i$ to be the first natural number $n > N_{i - 1}$ such that all $n \geq N_i$ satisfy $d(p_{\varphi_i(n)}, q_i) < \dfrac{1}{2^{i+1}}.$ Let $\psi_i = \varphi_i \circ \ldots \circ \varphi_1$ and $M_i = \varphi_{i-1} \circ \ldots \varphi_1(N_i),$ with $M_1 = N_1.$ Then $M_i \geq N_i$ and $\psi_i(N_i) = \varphi_i(M_i).$ Therefore, for all $i,$ $$d(p_{\psi_i(N_i)}, q) \leq d(p_{\varphi_i(M_i)}, q_i) + d(q_i, q) \leq \dfrac{1}{2^i}.$$
Thus, the subsequence $\left( p_{\psi_i(N_i)} \right)_i$ of $(p_n)$ converges to $q.$ A fortiori, $q \in E'.$ Q.E.D.
