As far as I can tell this involves little to no complex analysis. Also, I'm assuming you want $S$ to be open in the upper half plane, not necessarily in all of $\mathbb{C}$.
To construct $S$, consider any $[a,b] \times [0, R] \subseteq \mathbb{H}$ (that is, $\{z: a \leqslant Re z \leqslant b, 0 \leqslant Im z \leqslant R \}$, for $R >> 0$, by compactness and hence uniform continuity there exists a subset $S_{a,b}$ of the form $[a,b] \times [0, \epsilon)$ on which $f$ is nonvanishing.
Now taking $[a,b]$ to be say $\ldots [-1,0], [0,1], [1,2], \ldots$, and patching your $S_{a,b}$ together nicely at the endpoints gives $S$.
$S$ is clearly simply connected (it deformation retracts onto the real line via "squishing").
Lastly, to construct $\psi$, we have $\Omega := S - \mathbb{R}$ is open and simply connected, $f$ is nonvanishing on $\Omega$, and hence $f = e^{g(z)}$ for $g$ holomorphic. Take $\psi := g/i$.
If this is unclear (or wrong!) lemme know.
Your idea is correct: You start by defining the extended function
$$\tilde{f}:\mathbb{C}\to \mathbb{C},\quad \tilde{f}(z):=\begin{cases} f(z) & z\in\overline{\mathbb{H}}\\ \overline{f(z)} & z\in \mathbb{C}\setminus \overline{\mathbb{H}} \end{cases}.$$
By the Schwarz reflection principle this function is holomorphic on the whole of $\mathbb{C}$, i.e. entire.
But this entire function again satisfies $\Re(\tilde{f})\ge 0$. Now the function $$z\mapsto e^{-\tilde{f}(z)}$$
is again entire and is bounded, as
$$|e^{-\tilde{f}(z)}|=|e^{-\Re(\tilde{f}(z))}|\leq 1.$$
Thus by Liouville $e^{-\tilde{f}(z)}$ is constant. But this implies that $\tilde{f}(z)$ is constant and hence $f$ itself is constant.
Best Answer
One way this could be approached is to first let $G(z)=F(z+i), \Im z \ge -1$ so $G$ is analytic and bounded on $\{z|Im(z)\geq -1\}$ and let $f(w)=G(\frac{w+1}{i(1-w)}), |w|<1$ Then $f$ is analytic in the open unit disc, continuous and bounded on the unit circle except possibly at $1$ where it has bounded non-tangential limits, so $f \in H^{\infty}(\mathbb D)$
Since $dx/(x^2+1)=-2d\theta, \infty \to 1=e^{i0}, 0 \to -1=e^{i\pi}$, we have:
$\int_0^\infty \dfrac{\log(|F(x+i)|)}{1+x^2}dx=\int_0^\infty \dfrac{\log(|G(x)|)}{1+x^2}dx=2\int_0^{\pi}\log |f(e^{i\theta})|d\theta$
so we need to prove that if $f \in H^{\infty}(\mathbb D)$ and $\int_0^{\pi}\log |f(e^{i\theta})|d\theta=-\infty$, $f$ is identically zero.
This result is classic and holds with $[0,\pi]$ replaced by any set of non-zero Lebesgue measure on the unit circle by first noting that since $|f| \le M, \log |f| \le \log M$ so
$\int_0^{2\pi}\log |f(e^{i\theta})|d\theta \le \pi \log M+\int_0^{\pi}\log |f(e^{i\theta})|d\theta =-\infty$
Then since if we assume $f$ non-identical zero and $f(0)=0$ replacing $f(w)$ by $f(w)/w^n$ where $n \ge 1$ is the (finite) order of the zero of $f$ at $0$, nothing changes on the boundary as absolute values go ($|w|=1$!) there), we can assume $f(0) \ne 0$ so Jensen theorem shows that $\log |f(0)|+\sum_{|z_k|<r}\log(r/|z_k|)=1/2\pi\int_0^{2\pi}\log |f(re^{i\theta})|d\theta, z_k$ the (finitely many) roots of $f$ in $|z|<r<1$ so since $\log(r/|z_k|)>0, |z_k|<r$, we get
$2\pi \log |f(0)| \le \int_0^{2\pi}\log |f(re^{i\theta})|d\theta=\int_0^{2\pi}(\log^+ |f(re^{i\theta})|d\theta-\log^- |f(re^{i\theta})|)d\theta$ and using the boundness of $f$ we get:
$2\pi \log |f(0)| \le \int_0^{2\pi}-\log^- |f(re^{i\theta})|d\theta +2\pi \log M$
Now we can apply Fatou's lemma and get:
$\liminf_{r \to 1}\int_0^{2\pi}\log^- |f(re^{i\theta})|d\theta \ge \int_0^{2\pi}\liminf_{r \to 1}\log^- |f(re^{i\theta})|d\theta=\int_0^{2\pi}\log^- |f(e^{i\theta})|d\theta $ which gives:
$\int_0^{2\pi}\log|f(e^{i\theta})|d\theta \ge -\int_0^{2\pi}\log^-|f(e^{i\theta})|d\theta \ge (\limsup_{r \to 1})\int_0^{2\pi}-\log^- |f(re^{i\theta})|d\theta \ge 2\pi (\log |f(0)|-\log |M|) > -\infty$ so we get the required contradiction with $f$ (hence $G,H$) not identically zero and done!