How to show that the modulus of $\frac{z-w}{1-\bar{z}w}$ is always $1$?

You can do a no thinking calculation. Let $\displaystyle u=\frac{z-w}{1-\bar{z}w}$. We calculate $u\bar{u}$. Note that $\displaystyle \bar{u}=\frac{\bar{z}-\bar{w}}{1-z\bar{w}}$. Thus $$u\bar{u}=\frac{z-w}{1-\bar{z}w}\cdot\frac{\bar{z}-\bar{w}}{1-z\bar{w}}=\frac{(z-w)(\bar{z}-\bar{w})}{(1-\bar{z}w)(1-z\bar{w})}.$$ Expand the denominator. We get $$(1-\bar{z}w)(1-z\bar{w})=1-\bar{z}w-z\bar{w}+z\bar{z}w\bar{w}=1-\bar{z}w-z\bar{w}+z\bar{z}.$$ Expand the numerator. We get the same thing.

Another way (or hindsight is $20$-$20$): In $\displaystyle \frac{z-w}{1-\bar{z}w}$, replace the $1$ by $w\bar{w}$. We get $\displaystyle \frac{1}{w}\cdot\frac{z-w}{\bar{w}-\bar{z}}$. But $\displaystyle \frac{1}{w}$ has norm $1$, as does $\displaystyle \frac{z-w}{\bar{w}-\bar{z}}$.

Since $|w|=1$, $\bar w=\frac 1w$ hence $$\frac{z-w}{1-\bar z w}=\frac{z-w}{1-\bar z\frac 1{\bar w}}=\bar w\frac{z-w}{\bar w-\bar z},$$ which gives the result.