I realize this is probably in bad form to answer my own question, but the one comment I thought helped me solve the peice of the problem I was missing, is as I said, a comment, and not something which allows me to close this question.
Part $(a):$
Let $f(z)$ be an analytic function on a domain $D$ that has no zeroes on $D$. Suppose $|f(z)|$ attains a minimum on $D$ at $z_0$, denoted $M$. Since $f(z)$ has no zeroes on $D$, and is analytic, $g(z)=1/f(z)$ is analytic. It follows that $|g(z)|=|1/f(z)|\le 1/M$ for all $z\in\mathbb{Z}$ and $|g(z_0)|=1/M$. So, since $g(z)$ is analytic, it is also harmonic, and thus by the Strict Maximum Principle, $g(z)$ is constant, implying $f(z)$ is constant.
Part $(b):$
Let $f(z)$ be an analytic function on a domain $D$ that has no zeroes on $D$ and a zero on $\partial D$ at $z_0$. Then $f(z_0)=0$ implying $|f(z_0)|=0$, which is neccesarily a minimum on $\partial D$ since $|f(z)|> 0$ for all $z\in D$ as $f(z)$ has no zeroes in $D$.
Let $f(z)$ be an analytic function on a domain $D$ that has no zeroes on $D$ and $\partial D$. Suppose that $D$ is bounded and connected and that $f(z)$ extends continuously to $\partial D$. Then $D\cup\partial D$ is compact, as it contains its boundary and is bounded. Let $g(z)=1/f(z)$, then since $f(z)$ has no zeroes, $g(z)$ is defined over all of $D\cup\partial D$, and since $f$ is continuous, $g$ is continuous and thus it attains a maximum, denoted $M$ at $z_0$. Then $g(z)=1/f(z)\le M$, implying $f(z)\ge 1/M$ for all $z\in\mathbb{Z}$, and since $g(z_0)=M$, it follows that $f(z_0)=1/M$. If $|f(z)|$ obtains a minimum in $D$, it follows from part $(a)$ that $f(z)$ is constant and thus obtains its minimum on $\partial$$D$, and thus $|f(z)|$ always obtains its minimum on $\partial D$.
1) If $f$ had infinitely many zeros in the disk, those would have a limit point. By continuity the value at such a limit point would be $0$ too. If $f$ has no zeros on the circle, the limit point is thus a non-isolated zero of the function. But an analytic function with a non-isolated zero is identically $0$.
2) If $f$ had no zeros in the disk and constant absolute value on the circle, it would indeed have have constant absolute value, and therefore would be constant. A function with zeros in the disk can have constant absolute value on the circle, the simplest example being $f(z) = z$.
3) After dividing by the finite Blaschke product, you get a function with no zeros in the disk, and therefore you can apply (2).
Best Answer
By the maximum modulus principle, $f$ takes its maximum modulus on the boundary. By the minimum modulus principle (which is just the maximum modulus principle applied to $1/f$, which requires that $f$ have no zeros), $f$ also takes its minimum modulus on the boundary.
If the modulus is constant on the boundary, then the minimum modulus and the maximum modulus, both lying on the boundary, must be equal. Hence the modulus is constant on all of $D$ including the interior.
And if $|f|$ is constant on all of $D$, say $|f|(D)=\{K\}$, then the image of $D$ under $f$ lies inside the circle $\{e^{iθ}K\}.$ A circle which has empty interior in $\mathbb{C},$ so is not open.
But the open mapping theorem states that if a function $f$ is not constant, it must be an open map, i.e. it must send any open subset of $\mathbb{C}$ to an open subset.
Finally, by contraposition, since $f(D)\subseteq \{e^{iθ}K\}$ is not open, $f$ must be constant.