Is a map that preserves the hyperbolic distance biholomorphic? Let $\lVert z \rVert_w = \frac{|z|}{1 - |w|^2}$  be the hyperbolic distance in $\mathbb{D}$, and let the hyperbolic metric be $d(z, w) = \inf_\gamma \int_0^1 \lVert \gamma'(t) \rVert_{\gamma(t)} \, dt$. Automorphisms of $\mathbb{D}$ preserve this metric. I'm looking to prove the converse, that is, if $f \colon \mathbb{D} \to \mathbb{D}$ preserves $d$ then it is a biholomorphism from $\mathbb{D}$ to itself, or $\overline{f}$ is.
If $f$ is assumed $C^1$ it is mostly not so hard - you can get that at every point either $f$ is holomorphic or its conjugate is by considering $\psi_{f(0)} \circ f$, which preserves the origin and must also preserve the ordinary absolute value, hence is isotropic hence either $f$ or $\overline{f}$ is holomorphic at the origin, and since $f$ was arbitrary this is true of all points. (And hopefully this gives that one of them is globally holomorphic! But I haven't figured out that part yet.)
If nothing is assumed of $f$, is the result even true? I see why $f$ must be continuous, but does preservation of the hyperbolic metric imply $C^1$-ness?

Yes, this is true. The proof is not specific to hyperbolic metric: one can argue the same way about the Euclidean metric on the plane, or in higher dimensions.  
Step 1: Compose  $f$ with a Möbius transformation that sends $f(0)$ to $0$. This reduces the problem to the case $f(0)=0$. 
Step 2: Since $f$ is an isometry, every circle around the origin is mapped to self. By composing $f$ with a rotation, we reduce the problem to the case $f(1/2)=1/2$. 
Step 3: Consider the (hyperbolic) circles with centers $0,1/2$ passing through $i/2$. Like any two distinct circles, they have at most two points in common: specifically, $i/2$ and $-i/2$. Hence, $f(i/2)$ must be one of these points. By composing $f$ with conjugation, the problem reduces to the case $f(i/2)=i/2$.
Step 4. The same reasoning as in step 3 shows that $f(z)$ is either $z$ or $\bar z$ for every $z$. And since $d(f(z), i/2) = d(z,i/2)$, it must be $f(z)=z$.
