It would be much simpler to prove the result considering the family obtained by composing with the Möbius transformation $$z \mapsto \frac{z-1}{z+1}$$ that maps the right half plane biholomorphically to the unit disk.
But well, let's look at what we got from considering $e^{-f}$. Without loss of generality, we can assume that the entire sequence $e^{-f_n}$ converges compactly to a nonzero function $g$.
As you observed, that does not yet guarantee that the sequence $f_n$ itself converges compactly to a holomorphic function. So let's fix some $z_0 \in \Omega$ and consider the sequence $f_n(z_0)$. Either the sequence converges to $\infty$, or we can extract a subsequence converging to a complex number.
Consider first the case where we can extract a subsequence converging to a complex number. Without loss of generality, assume the entire sequence converges to $w_0 \in \mathbb{C}$. In a neighbourhood of $e^{-w_0}$, there is a branch of the logarithm with $\log e^{-w_0} = -w_0$ defined.
Then $f_n$ converges uniformly to $\log g$ in a neighbourhood of $z_0$.
If $f_n(z_0) \to \infty$, then, taking a branch of the logarithm in a neighbourhood of $g(z_0)$, we obtain a sequence $k_n$ of integers with $\lvert k_n\rvert \to \infty$ and $f_n(z_0) - 2\pi i k_n \to \log g(z_0)$. Thus the sequence $f_n - 2\pi i k_n$ converges uniformly to a holomorphic function in a neighbourhood of $z_0$, and since $\lvert k_n\rvert \to \infty$, the sequence $f_n$ itself converges uniformly to $\infty$ in a neighbourhood of $z_0$.
It remains to see that the uniform convergence to either a holomorphic function or $\infty$ extends (as locally uniform convergence) to all of $\Omega$.
Let $A = \{z \in \Omega : f_n(z) \to \infty\}$ and $B = \{z \in \Omega : f_n(z) \text{ is bounded}\}$ and $C = \Omega \setminus (A\cup B)$.
The argument above shows that all, $A$, $B$ and $C$ are open, and they are disjoint. Since $\Omega$ is connected, we have $\Omega = A$, $\Omega = B$, or $\Omega = C$. By having extracted the subsequence converging (to $\infty$ or $w_0$) at $z_0$, we have arranged that $z_0 \notin C$, hence $C = \varnothing$, so $\Omega = A$ if $f_n(z_0) \to \infty$, and $\Omega = B$ if $f_n(z_0) \to w_0$.
My question is now, is my suspicion grounded?
YES it is. In fact, there are, for any nonzero complex number $w$, exactly $k$ solutions to the equation $z^k=w$. This is because you can multiply any solution $z_1$ by $e^{\frac{2\pi}{k}i}$ (that is, rotate the solution around $0$ by an angle of $\frac{2\pi}{k}$) and still get a solution.
In your case, you could also have $$e^{\frac13(\ln|z| + i(\mathrm{Arg}(z) + \frac{1\cdot 2\pi}{3})}$$ or $$e^{\frac13(\ln|z| + i(\mathrm{Arg}(z) + \frac{2\cdot 2\pi}{3})}$$
and still get a solution.
Best Answer
Your argument is not correct. $\{1,2\}$ is compact and not a singleton but it has no limit points. You have to use the connectedness of the range of $h$ to say that there is a limit point.
[Any continuous map from a connected space into a discrete space is a constant].