How to show if any set $S$ is not closed, then there exists a sequence with no convergent subsequence I know that $S$ not closed implies there exists $x∈∂S$ such that $x∉S$. The definition of boundary then implies for every $ε>0$,
$B(ε,x)∩S≠∅$.
Since $x∉S$, it follows that
$∀ε>0$,$∃y∈S$ such that $y≠x$ and $|y−x|<ε$.
The idea is to use this assumption to show that we can find a sequence in $S$ that converges to $x$. Then every subsequence will also converge to $x$, and so (since $x∉S$) no subsequence can converge to a limit in $S$.
Now I wonder how to choose such sequence.
 A: You don't mention it, but this should be in the framework of a given metric space $(X,d)$.
If the set $S\subseteq X$ is not closed, take $x$ in the closure, but $x\notin S$.
For every positive integer $n$, we have $B(x,1/n)\cap S\ne\emptyset$. Choose $y_n\in B(x,1/n)\cap S$.
Then the sequence $(y_n)$ converges to $x$ (easy proof), so every subsequence thereof converges to $x$. Hence the sequence has no subsequence which is convergent to a point of $S$.
The additional remark above is necessary: if you take $S=(0,1)\subseteq\mathbb{R}$, then every sequence in $S$ has a subsequence that converges to some point of $\mathbb{R}$. Indeed, $S\subseteq[0,1]$, which is compact, so every sequence in $[0,1]$ (in particular in $S$) has a convergent subsequence.
A: If the set $S$ is not closed then there exists a limit point not in $S$.
That is the only thing that distinguishes a set that is not closed from an arbitrary set so if something can be done on any non-closed set that can't be done on any arbitrary set then it must have something to do with the limit point not in $S$.  It must.
And what is a limit point?  A limit point is a point $s$ so that every neighborhood of $s$ with radius $r$ no matter how small will contain a point $x_r \in S$ so that $d(x_r, s) < r$.
So we must use that to find a sequence that has no convergent subsequence.  Well we can find ever decreasing $r_i$ (for example, let $r_i = \frac 1i$) and let $x_i$ be any $x_i \in X$ so that $d(x_i, s) < r_i$. and that's it.  $\{x_i\}$ doesn't converge to any point is $S$ as it converges $s\not \in S$.
