Cauchy in $(\mathbb{R},d) \implies$ Cauchy in $(\mathbb{R},|\cdot|)$ Consider $\mathbb{R}$ with the metric $d$ such that $d(x,y) = \min\{1,|x − y|\}$ for all $x,y \in \mathbb{R}$. 
Prove that if $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence in $(\mathbb{R},d)$, then $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{R}$ with the standard metric.

My thoughts:
Since $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence in $(\mathbb{R},d)$ then given any $\epsilon>0$, there exists $N\in\mathbb{N}$ such that for all $n,m>N$ we have $d(x_n,x_m)=\min\{1,|x_n-x_m|\}=|x_n-x_m|<\epsilon$, since if $d(x_n,x_m)=1$ then $x_n,x_m$ would never be arbitrarily close. Hence the result.
Obviously this is a trvial solution, so I've no doubt misunderstood something! Any advice would be appreciated.
 A: You have the right idea, but your argument lacks formalism. A formal argument would go like this:
Let $\epsilon>0$. Let $\delta=\min\{\epsilon,1/2\}$. Since $(x_n)$ is Cauchy in $(\mathbb R,d)$, we know that there exists $N$ so that if $n,m\geq N$, then $$d(x_n,x_m)<\delta.$$ Since $\delta\leq 1/2<1$, we know that $$d(x_n,x_m)=|x_n-x_m|.$$ This means that $$|x_n-x_m|<\delta\leq\epsilon.$$
A: Note that $d(x,y) < 1$ iff $|x-y| < 1$.
Furthermore, if $d(x,y) < 1$ or $|x-y| <1$ then $d(x,y)=|x-y|$.
Suppose $x_n$ is $d$-Cauchy and let $\epsilon>0$. Let $\epsilon' = \min(\epsilon, {1 \over 2})$ and choose $N$ such that
$n,m \ge N$ we have $d(x_n,x_m) < \epsilon'$ Then $|x_n-x_m| = d(x_n,x_m) < \epsilon' \le \epsilon$. Hence $x_n$ is $|\cdot|$-Cauchy.
A: If $x_n$ is a Cauchy sequence in $\mathbb{R}$ then for any $\epsilon >0$ there is some $N_{\epsilon} \in \mathbb{N}$ (that will, in general, depend on $\epsilon$)  such that for all $n,m > N$, $d(x_n,x_m) < \epsilon$. Pick $\epsilon =1/2$, then for all $n,m > N_{1/2} $ we have $d(x_n , x_m) <1/2$ and hence $d(x_n,x_m) = |x_n - x_m|$. Hence the Cauchy property of a sequence holds in the new metric if and only if it holds in the standard one.
