Assume $f\colon \mathbb{D}\to\mathbb{R}^2$ is a harmonic map
and $x\notin f(\partial\mathbb{D})$. Is it true that $f^{-1}\{x\}$ is totally disconnected?


I hope that the answer is yes.
But actually I need a "yes", with a CAT(0) space instead of $\mathbb{R}^2$. 


So, I need a generalizable proof.


**P.S.** For CAT(0) target the answer is "no" --- if the target is a cone with large total angle then the tip of the cone might have a tree as an inverse image for harmonic map; this example is constructed in "Harmonic maps between flat surfaces with conical singularities" by Ernst Kuwert.



Harmonic maps are light
As your question operates with $f(\partial D)$, I assume that $f$ is continuous in
$\overline{D}$, though you do not mention this explicitly. Then the answer is yes, the zero set of $f$ is discrete (not just totally disconnected).


Suppose wlog that $0\not\in f(\partial D)$. Suppose by contradiction that zeros
of $f$ have an accumulation point $z^\*$ in $D$. Then, according to a theorem of Wilmshurst, there is an open analytic arc $\gamma$ which has $z^\*$ in its interior, and such that $f(z)=0$ on $\gamma$. Consider now this whole arc $\gamma$ (the maximal arc on which $f(z)=0$). It cannot reach the boundary $\partial D$ because
on $\partial D$ we have $|f(z)|>\delta>0$ by assumption. So $\gamma$ must contain a loop,
but then $f\equiv 0$ inside this loop and thus everywhere.


A. Wilmshurst, The valence of harmonic polynomials, PAMS 126, 7 (1998) 2077-2081,
Theorems 3, 4.


I don't know whether this argument generalizes to CAT(0) spaces because I do not know what is a harmonic mapping into a CAT(0) space.


