If $|z| < 1$, prove that $\Re \left(\frac{1}{1 - z} \right) > \frac{1}{2}$.

You need to use more than just $x,y<1$ for example with $x=y=\frac{3}{4}$ you obtain $\Re\left(\frac{1}{1-(x+iy)} \right)=\frac{2}{5} <\frac{1}{2}$.

Hint:

Using your computations: $$ \Re\left(\frac{1}{1-z}\right)=\frac{1-x}{(1-x)^2+y^2}=\frac{1-x}{x^2+y^2-2x+1}$$ but $x^2+y^2<1$ so: $$\frac{1-x}{x^2+y^2-2x+1}>\frac{1-x}{1-2x+1}=\frac{1}{2}$$

$$\Re \Big(\frac{1}{1-z}\Big)={1\over 2}\Big(\frac{1}{1-z}+\frac{1}{1-\overline{z}}\Big)$$ $$={1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+z\overline{z}}$$ $$>{1\over 2}\frac{1-\overline{z}+1-z}{1-z-\overline{z}+1}$$ $$={1\over 2}$$

I have also a pure geometric solution.

All points $z$ satisfying $|z|<1$ are ''inside'' of circle $\mathcal{C}$ with radius $r=1$ and center at $0$.

Now transformation $z\mapsto 1-z$ is reflection across $0$ followed by translation for $1$. So $\mathcal{C}$ ''moves'' to the right for $1$ and let this new circle be $\mathcal{C}'$.

Transformation $z\mapsto {1\over z}$ is inversion with center at $0$ with radius $1$ (i.e. circle $\mathcal{C}$). Since $0\in \mathcal{C}'$ and $\mathcal{C}'$ touches imaginary axis, $\mathcal{C}'$ maps to line $\ell$ which is parallel to imaginary axis and all the points ''inside'' of circle $\mathcal{C}'$ goes ''right'' to the $\ell$. Finally, since points of intersection of $\mathcal{C}$ and $\mathcal{C}'$, which have clearly real part ${1\over 2}$, are invariant, the line $\ell$ has equation $z={1\over 2}$ and thus conclusion.