Maximum of sum of finite modulus of analytic function.

Suppose by contradiction that the maximum of $f$ is an interior point $z_0$.

Write

$$f_i(z_0)= |f_i(z_0)| \omega_i \,,$$

with $\omega_i$ unit.

Let $g(z):= \sum_i \overline{\omega_i} f_i(z) \,.$

Then, for all $z \in D$ you have

$$| g(z)| = \left| \sum_i \overline{\omega_i} f_i(z) \right| \leq \sum_i \left| \omega_i f_i(z) \right| =f(z) \leq f(z_0)= g(z_0) = |g(z_0)|\,.$$

Now apply the maximum modulus principle to $g(z)$, and use the fact that if $g$ is constant then $|g(z)| \leq f(z) \leq g(z_0)$ implies $f$ is constant.

Yes: $|f_k|$ is a subharmonic function in $\Omega$, a sum of subharmonic functions is subharmonic, and a subharmonic function can't have a local maximum in a connected open set without being constant.