If the limit exists the set is closed 
Prove that if $p_n \to p$ in a metric space then the set of points $\{p,p_1,p_2, ...,\}$ are closed. 

A theorem in my book states that a set $S$ in a metric space is closed if and only if whenever $q_1,q_2,...$ is a sequence of points in $S$ that is convergent then the $\lim_{n\to \infty}q_n \in S$
What I did was try and prove it by contradiction. So suppose the set $S$ is not closed which will imply that $S^c$ is not open (where "not open" apparently doesn't mean "closed") then there exists a point in $S^c$; lets denote this point $p\in S^c$ such that there is an open ball $B(p,r)\in S^c$ ...
But I don't know how to finish it. 
 A: To complete your proof : You want to show that $S^c$ is open. Suppose not, then there exists a point $x \in S^c$ such that, for any $\epsilon > 0$, $B(x,\epsilon)$ is not a subset of $S^c$. In other words,
$$
B(x,\epsilon)\cap S \neq \emptyset \quad\forall \epsilon > 0
$$
For $\epsilon = 1/k$, there is a point $p_{n_k} \in S$ such that
$$
p_{n_k} \in B(x,1/k)
$$
Now the subsequence $\{p_{n_k}\}$ converges to $x$ (Why?)
However, $p_n \to p$, so $p_{n_k} \to p$ for any subsequence (Why?)
Hence, $x=p$ (Why?)
Thus, $p \in S^c$, which is a contradiction. Thus, $S^c$ must be open, and hence $S$ must be closed. Answering all the "why"s above should solve your problem :)
A: Let's prove that $p$ can be the only accumulation point of the set $S=\{p,p_1,\dotsc\}$
Suppose $q$ is an accumulation point of $S$ distinct from $p$, so $d(p,q)>0$. 
Set $\varepsilon=d(p,q)/2$; then there exists $k$ such that $p_n\in B(p,\varepsilon)$, for all $n>k$. This leaves only a finite number of elements of $S\cap B(q,\varepsilon)$, contradicting the fact that $q$ is an accumulation point. Indeed
$$
S\cap B(q,\varepsilon)\subseteq\{p_1,p_2,\dots,p_k\}
$$
because $B(q,\varepsilon)\cap B(p,\varepsilon)=\emptyset$.
Note also that, if $S$ has no accumulation points at all, it's closed.
Notation: $d$ is the metric on the space and $B(x,r)$ denotes the open ball with center $x$ and radius $r$.

Let's try it differently, by showing that the complementary set of $S$ is open, that is, it contains an open ball around any of its points.
Suppose $x\in X\setminus S$ (where $X$ is the whole space) and that every open ball $B(x,\varepsilon)$ intersects $S$.
If, for some $\varepsilon>0$, $p\notin B(x,\varepsilon)$, then $d(x,p)>\varepsilon$ and so
$$
B(x,\varepsilon)\cap B(p,r)=\emptyset
$$
where $r=d(x,p)-\varepsilon$. Since the sequence converges to $p$, there is an integer $k$ such that, for $n>k$, $p_n\in B(p,r)$. Therefore
$$
B(x,\varepsilon)\cap S\subseteq\{p_1,p_2,\dots,p_k\}
$$
and so, taking
$$
\delta=\min\{d(x,p_1),d(x,p_2),\dots,d(x,p_k)\}
$$
we have that $B(x,\delta)\cap S=\emptyset$, a contradiction.
Therefore, $p\in B(x,\varepsilon)$, for every $\epsilon>0$, which means that $x=p$, contradiction.
Thus, for any $x\in X\setminus S$, there is an open ball $B(x,\varepsilon)$ such that $B(x,\varepsilon)\cap S=\emptyset$, that is, $B(x,\varepsilon)\subseteq X\setminus S$, as we wanted to prove.

A different proof would be by noting that $S$ is compact, so closed in the space it lives in. But probably this is not a “simple” proof. ;-)
A: ... there is an open ball $B(p, r)$ that intersects $S$ for every value of $r$, from which you can construct a sequence in $S$ converging to a point in $S^c$...
But I don't know how to complete it!
A: I'd rather go for a direct proof: let $S={p,p_1,p_2,...}$ and $p*$ be an accumulation point of $S$. Let $\varepsilon>0$ be given. Since $p_n\rightarrow p$, there exists $n$ such that $$d(p_m,p)<\varepsilon/2$$ for all $m>n$. On the other hand, since $p*$ is an accumulation point, there exists $m>n$ such that $$d(p_m,p*)<\varepsilon/2.$$ 
So we have:$$d(p,p*)\leq d(p,p_m)+d(p_m,p*)<\varepsilon.$$ That implies that $d(p,p*)=0$ and so $p=p*$. But $p\in S$, so $S$ contains all its accumulation points.
Actually this is also a proof of the uniqueness of the limit. 
