I know how to prove that limits in a metric space are unique, but I'm trying to read an follow an alternative solutions from an instructor and I can't follow it. A paraphrase of his approach is as follows. The solution takes for granted the fact that a sequence $p_n \to p$ if and only if every open set $U$ containing $p$ contains $p_n$ for all but finitely many $n$. I'm bolding the sentence that gave me the most trouble.
Let $p_n \to p,p'$ with $p \neq p'$. Then for any $\epsilon > 0$, $B_{\epsilon} (p) \cap B_{\epsilon} (p')$ contains $p_n$ for all but finitely many $n$. But if $\epsilon = \frac{d(p,p')}{2}$, then $B_{\epsilon} (p) \cap B_{\epsilon} (p') = \emptyset$, which is a contradiction.
I don't understand why the bolded sentence is true. An open ball is open, and an intersection of open sets is open, but the quoted result requires that the open set contain $p$; if that's satisfied, then it contains $p_n$ for all but finitely many $n$. But it isn't true in general that for every positive $\epsilon$, the intersection of open balls contains $p_n$ for all but finitely many $n$, as we just showed. It seems to me that I would need to know a-priori that this open set contains either $p$ or $p'$.
What am I missing?
As $p_n\to p$ and $B_{\epsilon} (p) $ is an open set containing $p$ , it must contain all but finitely many terms of $(p_n) $ and similarly $B_{\epsilon} (p') $
$B_{\epsilon} (p) $ contains all but finitely many terms of $ p_n$ means $\exists k_1\in\Bbb{N}$ such that $ p_n\in B_{\epsilon} (p) , \forall n\ge k_1$
Again $B_{\epsilon} (p') $ contains all but finitely many terms of $ p_n$ . Hence $\exists k_2\in\Bbb{N}$ such that$ {p'}_n\in B_{\epsilon} (p') , \forall n\ge k_2$
Then $ p_n\in B_{\epsilon} (p) \cap B_{\epsilon} (p') ,\forall n\ge \max\{k_1, k_2\}$
Hence $ B_{\epsilon} (p) \cap B_{\epsilon} (p')$ contains all but finitely many terms of $(p_n) $.