harmonic function question Let $u$ and $v$ be real-valued harmonic functions on $U=\{z:|z|<1\}$.  Let $A=\{z\in U:u(z)=v(z)\}$.  Suppose $A$ contains a nonempty open set.  Prove $A=U$.
Here is what I have so far: Let $h=u-v$.  Then $h$ is harmonic.  Let $X$ be the set of all $z$ such that $h(z)=0$ in some open neighborhood of $z$.  By our assumptions on $A$, $X$ is not empty.  Let $z\in X$.  Then $h(z)=0$ on some open set $V$ containing $z$.  If $x\in V$, then $h(w)=0$ in some open set containing $x$, namely $V$.  So $X$ is open.  
I want to show $X$ is also closed but I am having trouble doing so.  Any suggestions:

Each real harmonic function $h$ on a simply connected domain defines unique up to the constant holomorphic function $f\in\mathcal{O}(U)$ such that 
$$
\mathrm{Im}(f)=h
$$ 
$$
\mathrm{Re}(f)=
\int\limits_{(x_0,y_0)}^{(x,y)}\left(\frac{\partial h}{\partial y}dx-\frac{\partial h}{\partial x}dy\right)+C
$$
If $h=0$ on some ball $B\subset A$, then respective function $f=C$ on $B$. Since $A$ is open, by uniqueness principle $f=C$ on $U$. Hence $h=\mathrm{Im}(f)=0$ on $U$.
