Uniform convergence of real part of holomorphic functions on compact sets The following is exercise 11.8 in Rudin's Real and Complex Analysis:

Suppose $\Omega$ is a region, $f_n \in H(\Omega)$ for $n = 1, 2, 3, \ldots$. $u_n$ is the real part of $f_n$. $\{u_n\}$ converges uniformly on compact subsets of $\Omega$, and $\{f_n(z)\}$ converges for at least one $z \in \Omega$. Prove that then $\{f_n\}$ converges uniformly on compact subsets of $\Omega$.

My thoughts: By Harnack's theorem, the limit $u$ of $\{u_n\}$ is harmonic. Thus it's the real part of a holomorphic function $f$ on $\Omega$ defined up to an imaginary constant. Using the limit of $\{f_n(z)\}$, we can find this constant. Thus $f$ is well-defined.
What's left is to show that the imaginary part of $f_n$ converges to that of $f$. I suspect I need to use the Cauchy-Riemann equations for this, but I cannot apply the familiar uniform convergence theorems with partial derivatives. What should I do?
Note: A region is a connected open subset of the complex plain.
 A: The following inequality helps: if $D$ is a disk centered at  $z_0$, $f$ is holomorphic in  $D$ and continuous on $\overline D$, then
$$\sup_{D'} |f|\le |\operatorname{Im}f(z_0)| + 3\max_{\partial D}|\operatorname{Re}f|, \tag1$$
where $D'$ is the disk concentric with $D$ and half the radius.
Proof of (1) is immediate from the Schwarz integral formula:
$$f(z)= i\operatorname{Im}f(z_0)+\frac{1}{2\pi i}\int_{\partial D}\frac{\zeta+z-2z_0}{\zeta-z} \,\operatorname{Re}f(\zeta)\,\frac{d\zeta}{\zeta-z_0}, \tag2$$
where $$\left|\frac{\zeta+z-2z_0}{\zeta-z}\right|\le 3,\qquad \zeta\in\partial D,\quad  z\in D'. \tag3$$
Returning to the problem at hand, an application of (1) to the differences $f_n-f_m$ yields uniform convergence of $f_n$ on any disk $D$ centered at $z_0$ and compactly contained in $\Omega$. Now (1) can be applied to  disks whose centers lie in the set in which uniform convergence is already known. Repeating in this fashion, we can cover any compact subset of $\Omega$ in finitely many steps.
A: I'm not entirely sure about this answer but I'll give it a try and people can point out any mistakes.
Let $z_{0}=x_{0}+iy_{0}$ be the point in $\Omega$ where the sequence $(f_{n})$ converges. We write $f_{n}(z)=2u_{n}(\frac{z+\bar{z_{0}}}{2},\frac{z-\bar{z_{0}}}{2i})-u_{n}(x_{0},y_{0})+ic_{n}$ where $c_{n}\in\mathbb{R}$. Letting $z=z_{0}$, we see that $c_{n}=\mathrm{Im}f_{n}(z_{0})$. Then uniform convergence of $(u_{n})$ on compact subsets and convergence of $(f_{n}(z_{0}))$ imply uniform convergence of $(f_{n})$ on compact subsets.
