Uniform convergence of harmonic functions to $0$ on compact subsets Let $D \subset \mathbb{C}$ be an open, connected set and let $\{ u_n \}$ be a sequence of harmonic functions with $u_n: D \longrightarrow (0, \infty)$. Show that if $u_n(z_0) \rightarrow 0$ for some $z_0 \in D$, then $u_n \rightarrow 0$ uniformly on compact subsets of $D$.
If you could offer a hint or a helpful question to get me started I'd appreciate it. 
Progress: I now see why showing it for the unit disk suffices (any compact subset in $D$ can be covered by finitely many disks each which is contained in $D$). 
 A: I think you should use identity theorem for harmonic functions...
Let $D_1$ be a disc with a center at $z_0$, and $D_1\subseteq D$. For every $n$ $u_n$ has harmonic conjugate $v_n$. So we can define new sequence of analytic function $\lbrace f_n\rbrace$ on $D_1$. 
$f_n = u_n + iv_n$. 
Let $f = \lim_{n\to\infty}f_n$. 
From Morera`s theorem $f$ is analytic on $D_1$, if, of course $D_1$ is small enough (maybe this is not so obvious).
Now define $g = \exp(f)$ and apply minimum principle to $g$. $|g(z_0)| = 1$ on $D_1$, so $f$ is constance on $D_1$, so $Ref = 0$ on $D_1$. Now you can apply identity theorem for harmonic function. 
A: As already mentioned in the comments, one considers the case of a disk first: If $u(z_0) \to 0$ and $\overline{B(z_0,r)} \subset D$ then Harnack's inequality (see also Harnack's inequality) shows that $u_n(z) \to 0$ uniformly in $\overline{B(z_0,r)}$.
For the general case, define
$$
 A = \{ z \in D \mid u_n(z) \to 0 \} 
$$
and show that both $A$ and $D \setminus A$ are open. Since $D$ is connected, one of the sets must be empty. $z_0 \in A$ implies that $A = D$.
So every point in $D$ has a neighborhood on which $u_n(z) \to 0$ uniformly, and that is equivalent to uniform convergence to zero on every compact subset of $D$.
