Question about the proof of the fact that subsequential limits are closed 
Theorem: For any sequence $\{p_n\}$ in a metric space $X$, the set $E$ of all subsequential limits of $\{p_n\}$ is closed relative to $X$.

I am not sure I follow the(?) proof of this theorem in my book so I'll just rewrite it in my words.
Let $q$ be a limit point of $E$. Then if $q \in E$, we are done. $\color{red}{\text{To that end we want to show some $\{p_{n_i}\}$ converges to $q$}}$.
Choose $n_1 \in \mathbb N$ s.t. $p_{n_1}$ is an element of $\{p_n\}$ with $p_{n_1} \ne q.$ Set $\delta = d(p_{n_1}, q)$.
Assume we have $p_{n_1}, \ldots p_{n_i}$ with $n_1 < \ldots n_i$ s.t. $p_{n_k} \ne q$ and $d(p_{n_k}, q) < \frac{\delta}{k}$ for any $k \le i.$
Since $q$ is a limit point, every neighborhood of $q$ contains a point $q \ne x \in E$ meaning $x \in N_{\text{any radius}}(q)$. In particular, $x \in N_{\frac{\delta}{2(i+1)}}(q)$ meaning $d(x, q) < \frac{\delta}{2(i + 1)}$. Since $x$ is a limit of some subsequence of $\{p_n\}$, past some $N \in\mathbb N$, for all $\epsilon > 0$, each element of that subsequence is closer than a distance of $\epsilon$ to $x$. In particular, that is true for $\epsilon = \frac{\delta}{2(i + 1)}$. So, for $n_{i + 1} > n_i > N$, we have $d(p_{n_{i + 1}}, x) < \frac{\delta}{2(i + 1)}$. Then $d(p_{n_{i + 1}}, q) \le d(p_{n_{i + 1}}, x) + d(x, q) < 2\frac{\delta}{2(i + 1)} = \frac{\delta}{i + 1}$. Now we have a subsequence $\{p_{n_i}\}$ with $d(p_{n_i}, q) < \frac{\delta}{i}$ for any $i \ge 2.$ From here, it's not hard to show $p_{n_i} \to p.$
I have two questions. Is my understanding of the proof correct? Also, regarding the part in red of the proof above, why does $p_{n_i} \to q$ imply $q \in E$?
 A: Your proof is essentially correct, but I think some points could be improved. Here is a suggestion:

*

*Let $n^{(q)}_1 = 1$ and $\delta = d(p_1,q) + 1 > 0$. [It is irrelevant whether or not $p_1 \ne q$.]


*Assume we have $p_{n^{(q)}_1}, \ldots p_{n^{(q)}_i}$ with $n^{(q)}_1 < \ldots < n^{(q)}_i$ such that $d(p_{n^{(q)}_k}, q) < \frac{\delta}{k}$ for any $k \le i.$ [It is irrelevant whether or not $p_{n^{(q)}_k} \ne q$.] Since $q$ is a limit point, every neighborhood of $q$ contains a point $x \in E \setminus \{q\}$. [Again it is irrelevant whether or not $x \ne q$.] In particular, we find $x \in E$ such that $d(x, q) < \epsilon := \frac{\delta}{2(i + 1)}$. Since $x$ is the limit of some subsequence $(p_{n^{(x)}_r})$ of $(p_n)$, for some $R \in \mathbb N$ each element $p_{n^{(x)}_r}$ of that subsequence with $r \ge R$ is closer than $\epsilon$ to $x$. Pick $r(i) \ge R$ such that $n^{(x)}_{r(i)} > n^{(q)}_i$ and let $n^{(q)}_{i+1} = n^{(x)}_{r(i)}$. Then we have $d(p_{n^{(q)}_{i + 1}}, x) < \frac{\delta}{2(i + 1)}$ and $d(p^{(q)}_{n_{i + 1}}, q) \le d(p^{(q)}_{n_{i + 1}}, x) + d(x, q) < 2\frac{\delta}{2(i + 1)} = \frac{\delta}{i + 1}$.


*Now we have a subsequence $(p_{n^{(q)}_i})$ with $d(p_{n^{(q)}_i}, q) < \frac{\delta}{i}$ for all $i$. Therefore $p_{n^{(q)}_i} \to q$. Hence by definition $q \in E$.
