Let $f(z) = \frac{1-z}{|1-z|}$ with $z\in\mathbb{C}$ such that $|z|< 1$. I want to know whether $f(z)$ is holomorphic in $|z|<1.$
Write $\frac{1-z}{|1-z|}=e^{i\arg(1-z)}$, where $\arg$ denotes the principal argument (from $-\pi$ to $\pi$). Then $$f(z)=e^{i\arg(1-z)}.$$ I know $e^w$ is an entire function of $w$. So it suffices to see if $\arg(1-z)$ is holomorphic in $|z|<1.$ Is this true?
The argument function has real values. If you think of the Cauchy-Riemann equations, it follows that it must be constant if it is holomorphic.
Your function is also not holomorphic. Assume for a contradiction that it is. By the maximum modulus principle, since $|f(z)|$ is constant, $f(z)$ has to be constant. But your $f(z)$ is not constant.