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?