Proof verification: $x_n \rightarrow a$ and $x_n \rightarrow b$ then $a=b$ Let $(x_n)$ be a sequence in a metric space $S$. Prove: if $x_n \rightarrow a$ and $x_n \rightarrow b$ then $a=b$.
Assume $a \neq b$. Take two balls $B_r(a)$ and $B_r(b)$ with such an $r$ that $B_r(a) \cap B_r(b)=\emptyset$. Then WLOG assume there is a sequence $(x_n)$ that converges to $a$. Then for all $\varepsilon > 0$, there is $N \in \mathbb{N}$ so that $d(x_n,a) < \varepsilon$ for all $n \geq N$. Then for $\varepsilon < r$, all the terms of the sequence past the corresponding $N$ are in $B_r(a)$ and not in $B_r(b)$, so the sequence certainly does not converge to $b$.
Is it correct enough?

Here is a simpler argument.
Take $\varepsilon>0$.
Then $d(x_n,a) < \varepsilon$ and $d(x_n,b) < \varepsilon$ for all $n$ sufficiently large.
But then $d(a,b) \le d(a,x_n)+ d(x_n,b) = d(x_n,a)+ d(x_n,b) < 2\varepsilon$.
Since $\varepsilon$ is arbitrary, this can only happen if $d(a,b)=0$, which implies $a=b$.
