Here, we have

$\large w= i({{1-z}\over{1+z}})$

$\large\therefore w+wz=i-iz\:\:\:\therefore z(w+i)=i-w$

$ \therefore\:z=\dfrac{i-w}{i+w}$

But, w = u+iv

$\large\therefore\:z={{i-w}\over{i+w}}={{i-(u+iv)}\over{}i+(u+iv)}={{i(-v+1)-u}\over{i(v+1)+u}}$

Since |z|=1,we get $|{{i(-v+1)-u}\over{i(v+1)+u}}|=1$

$\therefore\:|i(-v+1)-u|=|i(v+1)+u|\\ \therefore\:{(1-v)^2}+u^2=(1+v)^2+u^2\\\therefore\:1-2v+v^2+u^2=1+2v+v^2+u^2\\\therefore\:-4v=0\:\:\:\therefore v=0$

Thus,as v = 0,w = u + iv =u (real part of w only),as a result ,the circle |z| = 1 is mapped onto the real axis in the w-plane.

Hence,proved that the transformation w = ${{i-iz}\over{1+z}}$maps the unit circle |z|=1 into real axis of w plane.