jmc

Let $z=x+yi,$ where $x$ and $y$ are real numbers. Then from the condition $|z|=1,$ $\sqrt{x^2+y^2}=1,$ so $x^2+y^2=1.$

Now, $$\begin{array}{rl}\frac{1}{1-z}& =\frac{1}{1-x-yi}\\ & =\frac{1-x+yi}{(1-x-yi)(1-x+yi)}\\ & =\frac{1-x+yi}{(1-x{)}^2+y^2}\\ & =\frac{1-x+yi}{1-2x+x^2+y^2}\\ & =\frac{1-x+yi}{2-2x}.\end{array}$$The real part of this complex number is $\frac{1-x}{2-2x}=\frac{1-x}{2(1-x)}=\boxed{\frac{1}{2}}.$