We know that if every Cauchy sequence is convergent we call the space is complete . We have a theorem states every finite dimensional space should be complete . Let consider $(\mathbb R,d)$ where $d(x,y)=|\arctan(x)-\arctan(y)|$ it is obvious that $\mathbb R$ is finite dimensional but this space is not complete. Where is the error?
As Qiaochu Yuan has pointed out, the theorem you are thinking about concerns finite dimensional normed spaces. There is not a similar theorem for metric spaces, because we have no vector space structure on metric spaces and no concept of magnitude of elements in metric spaces.
Remember that completeness is a property of metric spaces, but every normed space is a metric space, with metric $\rho(x,y)=\|x-y\|$. It is with regards to this metric that we talk about completeness of normed spaces.
In your particular example, you have given a metric on the set $\mathbb R$, but no norm, so you cannot apply the theorem you are thinking of (It would be a useful exercise for you to inspect the proof of the theorem to see where the normed structure is used.) To show that the metric space $(\mathbb R,d)$ is not complete we need to find a Cauchy sequence that does not converge in this metric space. There are many sequences that will do the job. The simplest is the sequence $(x_n)=(n)$. If you have not already done so, I leave it to you to show that $(x_n)$ is Cauchy, but does not converge in $(\mathbb R,d)$, but feel free to ask for hints if you get stuck.
In light of what I said earlier about the theorem concerning the completeness of finite dimensional vector spaces, the incompleteness of $(\mathbb R,d)$ thus means that the metric $d$ cannot come from a norm on $\mathbb R$. It might be interesting to see why not.