There is no reason why your $z_0$ should be the maximum of $f$ along the boundary of any circle you construct. The maximum modulus principle just says the maximum of $f$ on a disc occurs at the boundary. If $z_0$ is a point on the boundary of a disc $B$, there may be $z_1$ on the boundary of $B$ such that $f(z_1) > f(z_0)$. If that is true then, of course, you may and will find $z_2$ in the interior of $B$ with $f(z_2) > f(z_0)$ but with $f(z_1) > f(z_2)$. There is no way to construct an inductive contradiction as you suggest.
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 $\partial D$.
Best Answer
Since $h$ and in particular $\textrm{Re}(h)$ is continuous on the closed unit disc it follows that $\lim_{r \uparrow 1} \max_{|z|=r} |\textrm{Re}\;h(z)| = 0$. Now the maximum of $|\textrm{Re}\;h(z)|$ for $|z| \leq r$ is attained on the circle $|z|=r$. This follows from the maximum/minimum principle for $e^h$. This implies that this maximum is non-decreasing in $r$. Since its limit is $0$ it must be identically zero. Therefore $\textrm{Re}\;h=0$ on the open unit disc. Then again it follows from the maximum/minimum principle for $e^h$ that $h$ must be constant.