What is the difference between the usual distance metric on $S^1$ and $d(e^{i\theta_1}, e^{i \theta_2})=\vert \theta_1-\theta_2 \vert$, where $\theta_1,\theta_2 \in [0, 2\pi)$ on $S^1$, where $S^1 $ is unit circle.and how I prove compactness of $S^1$ under $d(e^{i\theta_1}, e^{i \theta_2})=\vert \theta_1-\theta_2 \vert$.
Pls help to understand.
The usual distance metric on $S^1$ takes $d(e^{i\theta_1},e^{i\theta_2})$ to be the length of the shorter arc on $S^1$ connecting the points $e^{i\theta_1}$ and $e^{i\theta_2}$. This metric is equivalent to the metric $d'$ on $S^1$ defined by $$d'(e^{i\theta_1},e^{i\theta_2}):=|e^{i\theta_1}-e^{i\theta_2}|\ ,$$ since it is easily seen that $$d'(z_1,z_2)=2\sin{d(z_1,z_2)\over2}\qquad(z_1,\>z_2\in S^1)\ .$$ It follows that $S^1$ is compact as well under $d$ as it is under $d'$: The set $S^1$ is a closed and bounded subset of ${\mathbb R}^2$ according to the topology defined by the euclidean metric in ${\mathbb R}^2$, and $d'$ is nothing else but the euclidean metric restricted to $S^1$.
On the other hand your setup is special insofar as you insist on $0\leq\theta<2\pi$ when addressing the points $z\in S^1$. As a consequence the sequence $n\mapsto z_n:=e^{-i/n}$ does not converge to $1$ (as it does in the "standard" topology), but actually diverges. Note that $d_{1256}(z_n,1)\to2\pi$ $(n\to\infty)$. Your $S^1$ is indeed not compact.