If $p_n\to p$ in a metric space, then the set of points $\{p,p_1, p_2, p_3, \ldots \}$ is closed I'm having trouble with limits and convergence, bear with me.

Prove that if $\lim_{n \to +\infty} p_n = p$ in a given metric space, then the set of points $\{p,p_1, p_2, p_3, \ldots \}$ is closed.

Attempted Proof
Suppose that $\lim_{n \to +\infty} p_n = p$. This means that for any $\epsilon>0$, there is an index $N$ such that $d(p_n,p) < \epsilon$ for any given $n>N$. So to prove let there be an open ball around the point $p$ with radius $\min\{\epsilon, d(p, p_1) , d(p, p_2), ....\}$. However, since there is a distance $d(p_n, p) < \epsilon$ there will be a point in the sequence that is in the open ball thus causing a contradiction. Thus the limit is closed.
Is this proof correct?
 A: For $r\notin\left\{ p,p_{1},p_{2},\cdots\right\} $ find $N$ such
that $n>N\Rightarrow d\left(p_{n},p\right)<\frac{1}{2}d\left(r,p\right)$.
If $n>N$ then $d\left(r,p\right)\leq d\left(r,p_{n}\right)+d\left(p_{n},p\right)<d\left(r,p_{n}\right)+\frac{1}{2}d\left(r,p\right)$
and consequently $d\left(r,p_{n}\right)>\frac{1}{2}d\left(r,p\right)$.
Now take $\epsilon=\min\left\{ \frac{1}{2}d\left(r,p\right),d\left(r,p_{1}\right),d\left(r,p_{2}\right),\dots,d\left(r,p_{N}\right)\right\} >0$.
Then $d(r,x)\geq\epsilon$ for each $x\in\left\{ p,p_{1},p_{2},\cdots\right\}$
A: The statement is actually true in any Hausdorff space $X$, and in some ways working in that context is easier, once you’ve learned the basics of non-metric topology: there’s less unnecessary clutter. (Specifically, you don’t have to worry about the details of the metric.) This is a non-metric version of the proof that you attempted and that drhab’s answer presents correctly.

Let $P=\{p\}\cup\{p_k:k\in\Bbb Z^+\}$; we’ll show that $X\setminus P$ is open by showing that each point of $X\setminus P$ has an open nbhd disjoint from $P$. Let $q\in X\setminus P$; since $X$ is Hausdorff, there are disjoint open sets $U$ and $V$ such that $p\in U$ and $q\in V$. Since $\langle p_k:k\in\Bbb Z^+\rangle\to p$, there is an $m\in\Bbb Z^+$ such that $p_k\in U$ for all $k\ge m$, so $V$ is an open nbhd of $q$ disjoint from all of $P$ except the finite set $\{p_k:k<m\}$. Let $P_0=\{p_k:k<m\}$; being finite, $P_0$ is closed, and its complement is therefore open. Let $W=V\cap(X\setminus P_0)$; $W$ is open, since it’s the intersection of two open sets, and $q\in W$. Finally,
$$\begin{align*}W\cap P&=(W\cap\{p_k:k\ge m\})\cup(W\cap P_0)\\
&\subseteq(V\cap\{p_k:k\ge m\})\cup\big((X\setminus P_0)\cap P_0\big)\\
&=\varnothing\;,
\end{align*}$$
so $W$ is an open nbhd of $q$ disjoint from $P$. Thus, $X\setminus P$ is open, and $P$ is closed. $\dashv$

