Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete?

Your comment is correct. You can only get the final equality by proving that $\tan\left(\frac{\alpha}{2}\right)$ and $\frac{1-\cos \alpha}{\sin\alpha}$ have the same sign.

But this is not complicated to prove. $\tan\left(\frac{\alpha}{2}\right)$ is positive if and only if $\frac{\alpha}{2} \in (k\pi, k\pi +\frac{\pi}{2})$. Like $\sin \alpha$ while $1- \cos \alpha$ is always non negative.