Uniqueness theorem for harmonic function So far in complex analysis books I have studied about Uniqueness theorem: If $f$ is analytic in a domain $D$ and if its set of zeroes has a limit point in $D$ then $f\equiv 0$ on $D$, I want to know is this result holds for harmonic functions?
In general harmonic functions are functions $u:\mathbb{R}^n\rightarrow \mathbb{R}$. If $0$ is a regular value of such a function (which is true for almost every real number) then $u^{-1}(0)$ is locally an $(n-1)$ dimensional submanifold. Actually this is true for every $C^1$-function, regardless whether it is harmonic or not. So the question makes only sense for $u:  \mathbb{R}\rightarrow \mathbb{R}$. In this case the equation is $u^{''}=0$ which implies $u$ is linear.

So the answer is yes, in one dimension, cause there the solution is explicitly known, no otherwise, since the zero set is usually locally a submanifold of dimension $\ge1$.
I don't think this problem is true at all. 

The real or imaginary part of an analytic function is always an harmonic function. 

$f:\mathbb{C}\rightarrow\mathbb{C}:z\mapsto z$ is an analytic function thus $g:\mathbb{R^2}\rightarrow\mathbb{R}:(x,y) \mapsto \operatorname{Im}(g(x+iy))=y$ is a harmonic function. $g(x,y)=0$ for every point on the real line (points with $y=0$). But $g$ is not the zero function on any open subset of the complex plane (including the unit disc). 

Gotta hate harmonic functions.