Show that $(0,1)$ is completely metrizable

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Here, under the Section 'Examples ', $(0,1)$ is not complete with its usual metric inherited from $\mathbb{R}$, but it is completely metrizable since it is homeomorphic to $\mathbb{R}$.

To show $(0,1)$ is not complete with its usual metric, define a sequence such that $a_n=\frac{1}{n}$ for all $n \in \mathbb{N}$. Clearly the sequence is Cauchy but it does not have a limit in $(0,1)$, hence $(0,1)$ is not complete.

I'm having trouble to prove the open set is completely metrizable by defintion. I don't know how to construct such a metric. Can someone give some hints?

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HINT: Find a homeomorphism $f:(0,1)\to\Bbb R$, and define a metric $d$ on $(0,1)$ by

$$d(x,y)=|f(x)-f(y)|\;.$$

(Of course you’ll have to prove that this is a metric on $(0,1)$ and that it generates the right topology.)