Consider the union, over all rationals $p/q\in [0,1)$, written in lowest terms, of the line segments starting at $0$, having angle $2\pi p/q$ with the positive real axis, and having length $1/q$. Consider the union of this set with the unit circle and call this set $K$. Then $K$ is a compact, locally connected subset of the Riemann sphere (consisting of the unit circle, the interval [0,1] and countably many "spikes" protruding from zero. Let $U$ be the bounded complementary component of $K$, so $U$ is a simply-connected bounded subset of the plane, with locally connected boundary.
Let $\phi:\mathbb{D}\to U$ be a Riemann map, i.e. a conformal isomorphism between the unit disk and the domain $U$. By the Carathéodory-Torhorst theorem, the map $\phi$ extends continuously to the unit circle. This extension will have uncountably many zeros (one for each "access" to zero from the complement of $K$), but of course the map $\phi$ itself has no zeros.
To get an example of a holomorphic function that has infinitely many zeros, extends continuously to the boundary but has only one zero there (the minimum possible due to continuity) is very easy. For example, restrict the function \sin(z)/z to a horizontal half-strip surrounding the positive real axis, and whose boundary does not pass through any zeros. Precompose with a Riemann map taking the disk to this strip to get the desired map. (This is similar to J.J.'s example as above, but I've divided by z to ensure a continuous extension.)
Edit. It is worth noting that by the F. And M. Riesz theorem, the set of zeros on the boundary has zero one-dimensional Lebesgue measure.
Hints:
Note that (1),(3) are the negation of one another. This implies that for any give $f$, exactly one of them is the case (but, perhaps not that different examples cannot admit different cases).
Restate the context: $f$ is analytic on $\Bbb D$ and $1\in S^1\cap\mathrm{Rng}(f)$. We're asked whether $|S^1\cap\mathrm{Rng}(f)|$ is finite or infinite.
Added I think it's safe to elaborate on my earlier hints. Truth be told, though, I haven't noticed that the answer should best rely on the maximum modulus principle. In particular, I based my remarks on the open mapping theorem (which the maximum modulus principle can be viewed as a special case of). Indeed, if the range of $f$ contains $1$, it also contains some neighborhood thereof, which intersects a segment of the unit circle $S^1$. Since all these points must have distinct preimages, $f$ maps to $S^1$ on infinitely (uncountably) many points of $\Bbb D$.
Best Answer
Write $f=u+iv.$ Then $w(t) = u(e^{it})^2 + v(e^{it})^2$ is a real analytic function on $\mathbb R.$ We are given that $w(t)=1$ for infinitely many $t\in [0,2\pi].$ Thus $w=1$ on set with limit point in $[0,2\pi].$ By the identity principle for real analytic functions, $w\equiv 1.$ This implies $|f(e^{it})|\equiv 1.$ A well known exercise then shows $f=cB_1\cdots B_n,$ where $c$ is a constant of modulus $1,$ and the $B_k$ are Blaschke factors.