I want to prove that $u^{-1}(u(x_0))$ is infinite for dimension $N\ge 2$, for $x_0\in \Omega\subset\mathbb{R}^N$ and $u$ harmonic. I've found $u$ harmonic then $u^{-1}\{u(x_0)\}$ is infinite for $N\ge 2$ but I think I have an idea
By the maximum principle, both the maximum and the minimum of $u$ are attained at $\partial \Omega$. So construct a ball around $x_0$. $x_1$ and $x_3$ are the maximum and minimum points of $u$ in the border of the ball. By the intermediate value theorem and using the fact that the ball is connected, the image of $u$ around a path that passes through $x_1$ and $x_3$ is an interval. By the intermediate value theorem, there exists a point $x_2$ such that $u(x_1)<u(x_2)=u(x_0)<u(x_3)$. Since I can take infinitely many paths (for $N\ge 2$), the result follows.
Is it ok?
Take a closed ball $\overline B(x_0,r) \subset \Omega.$ Let $S_r=\partial \overline B(x_0,r).$ Then by the max/min principle, the minimum $m$ and maximum $M$ of $u$ over $\overline B(x_0,r)$ occur on $S_r.$ Now $S_r$ is connected. It follows that $u(S_r)$ is connected, hence is an interval, and therefore equals $[m,M].$ Since $m\le u(x_0)\le M,$ we have $u(x_0) = u(x)$ for some $x\in S_r.$
Now there are uncountably many $r$ such that $\overline B(x_0,r)\subset \Omega.$ For each one, the above shows there is $x_r \in S_r$ such that $u(x_r)=u(x_0).$ It follows that $u^{-1}(\{u(x_0)\})$ is not just infinite, but uncountable.