Let $S = \mathbb{R}\setminus \{0\}.$ Construct a metric $\rho$ on $S$ such that
(1) $(S,\rho)$ is a complete metric space and
(2) for any sequence $\{s_n\}$ in $S$ and $s \in S,$ the $\lim_{n\to\infty} \rho(s_n,s) = 0$ if and only if $x_n \to x$ in $(S,\rho).$
Prove that $\rho$ satisfies both properties.
First thought was to try the discrete metric, but this violates (2). My next idea was to use $\rho(x,y) = \frac{|x-y|}{1+|x-y|},$ but I have been unable to show that this makes $(S,\rho)$ complete. Any insight/suggestions?
HINT: For a complete metric that generates the usual topology, try $$d(x,y)=|x-y|+\left|\frac1x-\frac1y\right|\;.$$
Added: Suppose that $\sigma=\langle x_n:n\in\Bbb N\rangle$ is $d$-Cauchy. Let $\epsilon>0$; there is an $m_\epsilon\in\Bbb N$ such that $d(x_k,x_\ell)<\epsilon$ whenever $k,\ell\ge m_\epsilon$. But then $|x_k-x_\ell|\le d(x_k,x_\ell)<\epsilon$ whenever $k,\ell\ge m_\epsilon$, so $\sigma$ is Cauchy in the usual metric on $\Bbb R$. That metric is complete, so $\sigma$ converges to some $x\in\Bbb R$ with respect to the usual metric. To finish the argument, we must show that $x\ne 0$ and that $\sigma$ converges to $x$ with respect to $d$.
Suppose that $x=0$. By passing to a subsequence if necessary we may assume that the $x_n$ all have the same algebraic sign.
Show that $\lim\limits_{n\to\infty}\frac1{x_n}=\infty$ if the $x_n$ are positive, and $\lim\limits_{n\to\infty}\frac1{x_n}=-\infty$ if the $x_n$ are negative.
Show that either case contradicts the hypothesis that $\sigma$ is $d$-Cauchy, because for any $m\in\Bbb N$ and positive real number $\alpha$ there are $k,\ell\ge m$ such that $$\left|\frac1{x_k}-\frac1{x_\ell}\right|>\alpha\;.$$
This shows that $x\ne 0$, and therefore also that
$$\lim_{n\to\infty}\frac1{x_n}=\frac1x\;.$$
Now it should be straightforward to verify that $\lim\limits_{n\to\infty}d(x_n,x)=0$.