if a sequence $x_n$ in a metric space does not contain any convergent subsequence, then $x_n$ is closed in the metric space "If a sequence $(x_n: n\in\mathbb N)$ of a metric space does not contain any convergent subsequence, then the set $\{x_n: n\in\mathbb N\}$ is closed in the metric space"
I am trying to figure out which definitions make this proof the easiest. Since we are talking about sequences, closed should be best defined as "every sequence in the closed set converges to a point in the set." Now since the sequence $x_n$ does not contain any convergent subsequence, it must be unbounded by the Bolzano theorem.
Now can we say that the subset $x_n$ is closed because you cannot find a sequence in $x_n$ such that it converges to a point outside of $x_n$? (since no sequence even converges to anything at all)
 A: You cannot use Heine-Borel for any arbitrary metric space $X$.
We wish to show that the set $A = \{x_n \mid n \in \Bbb N\}$ is closed in $X$. Note that if $y$ is a limit point of $A$, then there exists a sequence of distinct points in $A$ converging to $y$. However, given a sequence of distinct points in $A$, you can find a corresponding subsequence of $(x_n)$.1 However, we know that $(x_n)$ has no convergent subsequences. Thus, $A$ has no limit points and is closed.
Note that I've also shown that no point of $A$ is a limit point.


*

*To elaborate more:
Suppose $(z_n)$ is a sequence of distinct points in $A$. Then, each $z_n$ is equal to some $x_{m(n)}$. Points being distinct implies that $n \mapsto m(n)$ is one-one.
Now, there are only finitely many naturals which are smaller than $m(1)$. Thus, we can find $n_2 > 1$ large enough so that $m(n_2) > m(1).$ A similar argument shows that we can find $n_3 > n_2$ large enough so that $m(n_3) > m(n_2)$ and so on.
Putting $n_1 = 1$, we see that $\left(z_{n_i}\right)_i$ is indeed a subsequence of $(x_n)_n$. Now, being a subsequence, $\left(z_{n_i}\right)_i$ has the same limit as $(z_n)_n$ and the argument goes through.

A: Let $S := \{ x_n \}$ be the set corresponding to the sequence.
$S$ is closed iff every limit point of $S$ is in $S$.
If there were some limit point $y$ of $S$ which was not in $S$, there would have to be a convergent sequence of points in $S$ converging to $y$, because we're living in a metric space: since $B(y, 1/n) \cap S$ is always nonempty for a limit point $y \in S^c$, we can pick $y_n \in B(y, 1/n) \cap S$ for every $n \geq 1$, and then $y_n \to y$ is a convergent subsequence of $\{ x_n \}$.
So if $\{ x_n \}$ has no convergent subsequences, then no such $y$ can exist. Done.
A: Let's have a sharper statement:
If a sequence $(x_n: n\in\mathbb N)$ of a metric space does not contain any subsequence convergent to any point different from any of $\ x_n,\ $ then the set $\{x_n: n\in\mathbb N\}$ is closed in that metric space.
Proof   For every point $\ y\ $ different from all $\ x_n\ $ there exists a radius $\ r_y>0\ $ such that open ball
$$ B_y\ :=\ \{z: d(y\ z)<r_y\} $$
does not contain any of $\ x_n.\ $ Then the union of such balls,
$$ \bigcup_y\, B_y $$
is an open set, and $\ \{x_n: n\in\mathbb N\} $ is closed because it is the complement of the said union.
