We have the set $\left \{z\in \mathbb{C}\mid z\neq 1, \ \left |\frac{z}{z-1}\right |\leq 1\right \}$
Let $z=x+yi$.
We have that \begin{align*}\frac{z}{z-1}&=\frac{x+yi}{(x-1)+yi}=\frac{\left (x+yi\right )\left ((x-1)-yi\right )}{\left ((x-1)+yi\right )\left ((x-1)-yi\right )} \\ & =\frac{x(x-1)+y^2-yi}{(x-1)^2+y^2} =\frac{x(x-1)+y^2}{(x-1)^2+y^2}+\frac{-y}{(x-1)^2+y^2}i\end{align*}
Then we get \begin{align*}\left |\frac{z}{z-1}\right |&=\sqrt{\left (\frac{x(x-1)+y^2}{(x-1)^2+y^2}\right )^2+\left (\frac{-y}{(x-1)^2+y^2}\right )^2} \\ & =\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4}{[(x-1)^2+y^2]^2}+\frac{y^2}{[(x-1)^2+y^2]^2}} \\ & = \sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}} \end{align*}
can we simplify that further?
Is not, we have to solve the inequality from that point. So we have from $\left |\frac{z}{z-1}\right |\leq 1$ the following:
\begin{align*}&\sqrt{\frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}}\leq 1 \\ & \Rightarrow 0\leq \frac{x^2(x-1)^2+2x(x-1)y^2+y^4+y^2}{[(x-1)^2+y^2]^2}\leq 1 \\ & \Rightarrow 0\leq x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq [(x-1)^2+y^2]^2 \end{align*}
From the second inequality we have \begin{align*}&x^2(x-1)^2+2x(x-1)y^2+y^4+y^2\leq (x-1)^4+2(x-1)^2y^2+y^4 \\ & \Rightarrow x^2(x-1)^2+2x(x-1)y^2+y^2\leq (x-1)^4+2(x-1)^2y^2\end{align*} How do we continue from here? I got stuck right now.
It's much easier to rearrange to $|z|\le|z-1|$. Geometrically, the distance of $z$ from $0$ is less than or equal to the distance of $z$ from $1$ in the complex plane, so the complex numbers $z$ satisfying this are those with $\Re(z)\le\frac{1}{2}$, or $x\leq\frac{1}{2}$.
Alternatively, you can write $z=x+iy$ in $|z|\le|z-1|$ and squaring both sides gives $x^2+y^2\le (x-1)^2+y^2$, which simplifies to $0\le 1-2x$, which has the same solution.