The set of all subsequential limits of a sequence is closed. 
Let $X$ be a metric space and $\{x_n\}$ be a sequence in $X$. Let $E$ be the set of all subsequential limits of $\{x_n\}$. Then $E$ is closed.

I have a idea of this proof.
Assume $E$ is a infinite set.
Let $p\in X-E$. Then $\exists r>0$ such that $B(p,r)\cap E=\emptyset$. The reason is if $p\in \bar E$, then there is some subsequential limit $y$ close to $p$, and there exists a subsequence $x_{n_k}$ converges to $y$. So, $d(x_{n_k},p)<d(x_{n_k},y)+d(y,p)<\varepsilon$ which is a contradiction.
So, $B(p,r)\subset X-E$ and $E$ is closed.

Allow me to give you an indication of a relation that holds in very general settings. Consider a first countable topological space $(X, \mathscr{T})$, in other words a space in which every point $x \in X$ admits a countable neighbourhood base.
Given an arbitrary family $x$ of objects (sets, ultimately) indexed by set $I$ and an arbitrary subset $J \subseteq I$, let us write $x_J\colon=\{x_i\}_{i \in J}$ for the set of components of indices $i \in J$ of the family $x$. For arbitrary natural number $m \in \mathbb{N}$, let us denote the unbounded natural interval of all natural numbers $n \geqslant m$ by $[m, \rightarrow)$.
Given an arbitrary sequence of points $x \in X^{\mathbb{N}}$, it is not difficult to show that the set of limits of subsequences of $x$ is equal to the set of points adherent to the sequence $x$, set which is by definition expressed as $\displaystyle\bigcap_{n \in \mathbb{N}}\overline{x_{[n, \rightarrow)}}$. As an intersection of closed subsets, this subset is itself closed.
Metric spaces are in particular first countable (for any point $x \in X$ the collection $\left\{\mathrm{B}_q(x)\right\}_{q \in \mathbb{Q_{+}^{\times}}}$ of open balls centered at $x$ and of rational radius is a neighbourhood base), so the above statement particularly applies.
