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.

A more direct but conceptual way is the following. Suppose $(x_n)\in X^{\mathbb{N}}$ is a sequence in the metric space $X$. Denote $L$ the set of all limits of the subsequences of $(x_n)$. Then
\begin{align}
L = \bigcap_{k\geqslant 0}\overline{\{x_n ~|~ n \geqslant k\}}
\end{align}
It is an intersection of closed subset, thus it is a closed subset.
This is because a point $l$ is in $L$ if and only if for every $\varepsilon >0$ and every $n \geqslant 0$, there exists $k \geqslant n$ such that $d(l,x_k) \leqslant \varepsilon$.
