Let $f_n:D\to\mathbb{C}$ be a sequence of holomorphic functions for which $|f_n(s)|^2$ converges uniformly, i.e. for $\varepsilon>0$ there always exists $N>0$ such that for all $n,m>N$ we have $||f_n(s)|^2-|f_m(s)|^2|<\varepsilon$ is true.
Does it follow that $f_n(s)$ is also uniformly convergent?
Assuming $||f_n(s)|+|f_m(s)||\neq 0$, then from my working so far I know that $$||f_n(s)|-|f_m(s)||<\frac{\varepsilon}{||f_n(s)|+|f_m(s)||}<\varepsilon,$$ but then I'm unsure if I can proceed...
No, it doesn't: Set $S:=[0,1]$ and $f_n(s):=\exp(i\pi\cdot s^n)$. Then $|f_n|^2$ is equal to $1$ everywhere on $S$, hence $|f_n|^2$ converges uniformly. $f_n$ converges pointwise to the function $$f(s):=1, s\in[0,1); f(1):=-1.$$ Since all $f_n$ are continuous but $f$ isn't, the convergence cannot be uniform.