If a set $S\subset\mathbb R$ is not closed, does it contain a convergent sequence with a limit outside of $S$? Suppose S is a subset of $\mathbb{R}$ and that S is not closed. Must it follow that there is a convergent sequence in S that converges to some l not in S?
 A: Yes.  If $S$ is not closed, it means that $S$ does not contain all of its limit points.  In particular, there exists some $x \in \Bbb R \setminus S$ such that for every $\epsilon > 0$, there is an $s \in S$ such that $|x-s|<\epsilon$.
Now, for each $n \in \Bbb N$, select an $s_n$ such that $|x - s_n| < 1/n$.  Consider the sequence $\{s_n\}_{n=1}^\infty$.
A: yes, if you take a point $x\in \bar{S}\backslash S$ and for $n\in\mathbb{N}$ the neighborhood $V_n(x)=]x-\frac{1}{n+1},x+\frac{1}{n+1}[$ of $x$ must continent a element $x_n$ of $S$ because $x\in \bar{S}$ so we have a sequence $(x_n)_n\subset S$ of elements of $S$ witch verify : 
$$
|x_n-x|\leq \frac{1}{n+1}
$$
so $\lim_{n\rightarrow\infty}x_n=x $
A: If $F$ is not closed $F\not=\overline F$, so there is some $x\in \overline F\setminus F$. For all the open balls around $x$ we have $B(x,r)\cap F\not=\varnothing$ (otherwise $B(x,r)\subset \bar{F}^c$). Let $r=1, \,1/2, \,1/3,\ldots$ and choose  elements $x_n\in B(x,1/n)\cap F$. S0 $x_n\to x$, $x_n\in F$ but $x\notin F$.

More generally:
For any topological space we have the following beautiful characterization: $F$ is closed iff for every net $x_i\to x$ in $S$ with $x_i\in F$ we have $x\in F$ (if you don't know what a net is, don't worry in a metric space, as the reals,  nets can be replaced by sequences)
Lemma: For any  $A\subset S$, $\overline A$  is the set of all $x\in S$ such that some net $x_i\to x$ with $x_i\in A$
Proof: Suppose $x\not\in  \overline A$ and $x_i\to x$ eventually for some $i$, $x_i\notin \overline A$. Let $x\in \overline A$ and $\mathcal{N}_x$ the filter of all the nbhd of $x$. Let $N\in \mathcal{N}_x$, so $N\cap A \not=\varnothing$ (if not, for an open $x\in U\subset N$, $U\subset S\setminus \overline A$ ). The set $\mathcal{N}_x$ is directed by ´"$\supset$" and choose $x_N\in N\cap A$. Thus the net $x_N$ converges to $x$. 
Proposition For a topological space $(S,\mathcal{T})$, $F$ is closed iff for every net $x_i\to x$ in $S$ with $x_i\in F$ we have $x\in F$
Proof: A set $F$ is closed iff $F=\overline F$ and for the above lemma the result follows.
