Suppose that $U$ is an open, connected subset of complex plane. Are there any non-constant harmonic functions on $U$ which have at least one closed isocurve?
My results so far:
If U is simply connected, this is impossible by maximum principle.
If U is not simply connected then we can construct harmonic functions which don't have a harmonic cojugate, i.e. harmonic $u$ for which there is no harmonic $v$ such that $u+iv$ is holomorphic. An example is $u(x,y)=\mathrm{Re} \log (x+i y)$ defined on $U = \mathbb C - \{0 \}$ (imaginary part is discontinuous on this set). Its isocurves are circles. From now I restrict attention to the case where this doesn't happen, so $u=\mathrm{Re} f(z)$ for some $f$ holomorphic in $U$.
If a closed isocurve $C$ exists, it contains two critical points of $f=u+iv$. That's because we have $$ 0 = \int_C \mathrm{d}z f'(z) = \int_C \mathrm d s \vec n \cdot \vec \nabla v, $$ where $\vec n$ is unit tangent vector to $C$ and $\mathrm d s$ is arc-length element. However it follows from Cauchy-Riemann equations and the fact that $u$ is constant on $C$ that $\vec \nabla v$ is parallel to $\vec n$. Therefore if $\vec \nabla v$ doesn't vanish anywhere on $C$ then $\vec n \cdot \vec \nabla v$ has constant sign on $C$, so we have a contradiction.
Examples are provided by functions $e^z-e^{\frac{1}{z}}$ as well as $z - \frac{1}{z}$. This was pointed out in comments and the question was never officially answered, so I decided to post this as community wiki.