Here's how I think the proof goes:
$(1)$ Choose an arbitrary limit point of the set $E^*$.
$(2)$ Choose a certain neighborhood of $q$ which contains some point $x$(which is one of infinitely many points).
$(3)$ Use the fact that some subsequence {$p_{n_i}$} converges to $x$.
$(4)$ Show that this subsequence converges to $q$ and hence that $E^*$ contains all its limit points and is closed.
Here are the questions:
$(1)$ The final conclusion is that the subsequence {$p_{n_i}$} converges to $q$, but didn't he also use the fact that it converges to $x$? Doesn't this imply that $q=x$?
$(2)$ The point $x$ chosen in the proof is by no means special so this argument can be applied to infinitely many points in $E^*$. Keeping in mind that we can do this for an arbitrarily large neighborhood of $q$, doesn't that show that $E^*$ contains only one point namely $q$ as the process of choosing arbitrarily large neighborhoods of $q$ and applying this argument to every point in them never terminates?
$(3)$ Why did he define $\delta$ to be the distance between $q$ and a specific point? couldn't he have chosen $\delta$ arbitrarily and still found infinitely many $x$'s on which he may apply the argument since $q$ is a limit point?
These questions didn't come up when I first read the proof, they occurred to me when I tried to prove it myself and the proof was an epic failure.
There's an issue with your understanding. The idea is to inductively/recursively build a subsequence $(p_{n_l})$ of $(p_n)$ converging to $q$. The point $x$ (which depends on $l$, so you can think of it as $x_l$ or something) is to help construct the next point $p_{n_l}$ with desirable properties, based on $p_{n_{l-1}}$.
Think of it this way: if $q$ is in the closure of $E^*$, then we can select a point $x \in E^*$ that's "really close" to $q$ (and yes, provided $q$ is not isolated in $E^*$, there will be infinitely many choices for $x$). But, since $q$ is a limit point of $x$, then points of the sequence $(p_n)$ become "really close" to $x$. An application of triangle inequality means that these points of the sequence $(p_n)$ are really close to $q$. The rest of the argument is just making sure that these points of $(p_n)$ line up to form a subsequence (i.e. ensuring that $n_l$ is a strictly increasing sequence of natural numbers).