I am trying to prove the following result.
Show that a sequence $p_1, p_2, p_3, \ldots$ of points in a metric space $X$ can have at most one limit.
The proof I have in mind is direct, but no other solution I've read, including in textbooks, use this approach. Many of them use contradiction and a lemma (that I'll write below), which makes me think there is an error in my approach or some reason I'm not seeing that my proof is inferior.
My approach:
Suppose $p_n \to p,p' \in X$. We will show that $p = p'$. Fix $\epsilon > 0$. As $p_n \to p$, we can find $N \in \mathbb{N}$ such that for every $n \geq N$, $d(p_n, p) < \frac{\epsilon}{2}$. Analogously, as $p_n \to p'$, we can find $N' \in \mathbb{N}$ so that for every $n \geq N'$, $d(p_n, p') < \frac{\epsilon}{2}$. For $n \geq N'' := \max(N,N')$, we then have $$ d(p,p') \leq d(p, p'') + d(p'', p') < 2\left(\frac{\epsilon}{2}\right) = \epsilon. $$ As $\epsilon > 0$ is arbitrary, we concle that $d(p,p') = 0$, so $p = p'$, and the limit is unique.
The approach by contradiction takes the following lemma for granted:
$p_n \to p$ if and only if every open set $U \owns pi$ contains $p_n$ for all but finitely many $n$.
The proof is then:
Suppose that $p_n \to p, p' \in X$ where $p \neq p'$, so $d(p,p') \neq 0$. Setting $\epsilon = \frac{d(p,p')}{2} > 0$, we then have that $B_{\epsilon} (p) \cap B_{\epsilon} (p')$ contains $p_n$ for all but finitely many $n$ by the lemma. This is a contradiction, as $B_{\epsilon} (p) \cap B_{\epsilon} (p')$ is in fact empty.
(The solution I'm looking at stops here, but I'll continue the line of argument to demonstrate this fact, which is not immediately obvious to me.)
Indeed, if $q \in B_{\epsilon} (p) \cap B_{\epsilon} (p')$, then $d(p,q) < \epsilon$ and $d(p',q) < \epsilon$, so $$ d(p,p') \leq d(p,q) + d(q,p') < 2\epsilon. $$ As $\epsilon = \frac{d(p,p')}{2}$, we have $d(p,p') = 2\epsilon$, so this implies $2\epsilon < 2 \epsilon$, which is a contradiction.
So my questions are:
(1) Are both of these proofs mathematically correct (especially the first)?
(2) Is one superior/stronger? I feel like the first approach might be stronger because I only had to work from the limit definition and didn't need to prove that lemma. I also tend to prefer direct proofs, wherever possible, to proofs by contradiction, and it's my understanding this is generally the preference in upper-level math courses.
(3) Is my understanding of why $B_{\epsilon} (p) \cap B_{\epsilon} (p')$ in the second proof correct, or is there a simpler reason?
(1) Yes, your first proof is correct, though $p''$ should be $p_{n}$ for some $n \geq N$.
(2) If you define stronger/weaker as the number of other assumptions beside the definitions required to prove a statement, then yes your proof is "stronger" than the second proof. However, I believe that as long as a proof is valid, its elegance is subjective.
(3) Yes, your argument is correct. Intuitively, as $\epsilon = 2\cdot d(p,p')$, if you draw $\epsilon$-discs at $p$ and $p'$, they will not intersect.