Let $f(z) = \sqrt{(z(z-1))}$, is this reasoning sound for finding the branch cut of $f$?
$z^{1/2}$ fails to be continuous across the positive real axis i.e on $\mathbb{R}_{\geq 0}$ whereas $(z-1)^{1/2}$ fails to be continuous across $\mathbb{R}_{\geq 1}$, so we take the intersection of these two sets and conclude that the branch cut for $f$ is $\mathbb{R}_{\geq 1}$.
I am not sure if this approach is correct.
Technically, for functions like $\sqrt z$, $\mathrm{Ln}(z)$ and $z^\alpha$, there are infinitely many branches. For example, the accepted, standard definition of $\mathrm{Ln}(z)$ from $\ln z$ is defined as follows: $$ \mathrm{Ln}(z)=\ln r+i\theta\quad,\quad -\pi<\theta\le\pi. $$ With this definition, the non-analytical branch of $\mathrm{Ln}(z)$ is $z=re^{i\pi}$ for $r\ge 0$. However, defining $\mathrm{Ln}(z)$ as $\ln r+i\theta$ with $-\pi+\theta_0<\theta\le\pi+\theta_0$, would leave us with a non-analytical branch of $z=re^{i\pi+i\theta_0}$ with $r\ge 0$. Same thing happens with $\sqrt{z}$. For example, we can define $$ \sqrt{z}=r^\frac{1}{2}\exp\left(i\frac{\theta}{2}\right)\quad,\quad -\pi<\theta\le \pi, $$where the non-analytical branch would again be $z=re^{i\pi}$ for $r\ge 0$. Back to your question, since $z(z-1)$ is complete, the non-analytical branch of $\sqrt{z(z-1)}$ is $z(z-1)=-r$ for $r\ge 0$, or $z\in \{(x,0):0\le x\le 1\}\cup \{(\frac{1}{2},y):y\in \Bbb R\}$.