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?
 A: It's mostly fine, but it's even nicer if you say "and take $r = d(b, a)/2 > 0$"; then your reader knows that $B_r(a) \cap B_r(b) = \emptyset$. Also, "WLOG assume there is a sequence $(x_n)$ that converges to $a$" is a bunch of fancy words that say nothing. Why? Because the sequence $a, a, \ldots,$ is a sequence that converges to $a$. There's no "loss of generality" here, and no need to assume anything. Furthermore, you can't assume anything about $(x_n)$ -- it's given in the hypotheses. 
So a rewrite might be:
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 $a \neq b$, and let $r = \frac{d(a, b)}{2}$. Because $a \ne b$, we have $r > 0$. Then $B_r(a) \cap B_r(b)=\emptyset$. 
By definition of convergence, for all $\varepsilon > 0$, there is $N \in \mathbb{N}$ so that $d(x_n,a) < \varepsilon$ for all $n \geq N$. In particular, for $\varepsilon = r$, there is such an $N$. That means that 
$$
d(x_n, a) < r
$$
for all $n > N$, 
and hence (because of the empty intersection of balls) that
$$
d(x_n, b) > r
$$
for all $n > N$. But that contradicts the definition of $x_n \to b$, for the choice $\varepsilon = r$. 
A: 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 $a=b$.
