Clarification on " the collection of subsequential limits of a sequence in a metric space form a closed subset"

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 $q\in E^*$.

Maybe the following approach is more inuitive: if $q\in \overline {E^*}$, then $B_q(1)$ contains a point $x\in E^*$ and hence a point $p_{n_1}$ of $(p_n)$ (why?). Suppose $n_i:1\le i\le j-1$ are chosen so that $p_{n_i}\in B_q(1/i)$. Then, $B_q(1/j)$ contains a point $y\in E^*$ and hence point $p_j\notin\{p_{n_1},\cdots,p_{j-1}\}.$ The induction therefore proceeds and we get a sequence $(p_{n_j})\to q.$