I was asked to construct a a metric on $(0,1)$ that induces the usual topology on it and that turns it into a complete metric space. The obvious choice is to define the metric as follows:
We have the bijection, $$ f:(0,1)\to\mathbb{R}, \ x\mapsto\tan\pi\left(x-\frac{1}{2}\right)$$ And we now define the metric as $$ d(x,y)=|f(x)-f(y)| $$ which makes $((0,1),d)$ complete, because $\mathbb{R}$ is complete in its usual metric, and it induces the standard topology because $f$ is a homeomorphism.
Now I have to check that the sequence $(x_n)$ with $x_n = 1/n$ is not Cauchy in this metric. And I'm unable to do that. Can I get some assistance?
Just use the fact that $\lim_{n\to\infty}\tan\left(\pi\left(\frac1n-\frac12\right)\right)=-\infty$. In particular, the limit does not exist in $\mathbb R$. Therefore, your sequence does not converge. Since your space is complete, your sequence cannot be a Cauchy sequence.