This is the approach I took to solve this but I got stuck.
Suppose$f=u+iv\in $ {$z\in \mathbb{C}| |z-2|=1$} and that $f$ is analytic on an open connected set.
Then we have that $(u-2)^2+v^2=1$. Since $f$ is analytic it satisfies that Cauchy-Riemann equations.
By partial differentiating the above equation with respect to $x$ and $y$ I got $$(u-2)u_x=-vv_x$$ $$(u-2)u_y=-vv_y$$ By multiplying these 2 equations and using the CR equations I finally got that $$u_xu_y=0$$ This is where I am stuck. If I can prove that $v_x,u_x=0$ then from that I can get that $f'(z)=u_x+iv_x=0\implies f $ is constant. I hope someone can help me out. Thanks
Let $g = f - 2$. Then $g$ is analytic and has constant modulus $|g| = 1$. So $g$ is constant.
Here is a short proof. Write $g = u + iv$. Then $u^2 + v^2$ is constant, implying that $uu_x + vv_x = 0$ and $uu_y + vv_y = 0$. The Cauchy-Riemann equations in turn imply that $uu_x - v u_y = 0$ and $uu_y + v u_x = 0$. Thus $$u^2 u_x - uvu_y = 0 \quad \text{and} \quad uvu_y + v^2 u_x = 0.$$ Add these to get $(u^2 + v^2) u_x = 0$. Since $u^2 + v^2 = 1$ you get $u_x = 0$ in the open set.
You can show in a similar manner that $u_y = v_x = v_y = 0$ in the open set. As long as it is connected, $u$ and $v$ are constant.