 Solution

Let $d=\mathrm{diam}(A)$. Unless $A$ is empty, its closure meets the real axis; otherwise a loop in $\mathbb{H}\setminus A$ surrounding $A$ could not contract, contrary to $\mathbb{H}\setminus A$ being a simply connected domain. Choose $x\in \overline{A}\cap \mathbb{R}$. Then $A\subseteq \{z\in \mathbb{H}:|z-x|\le d\}$.

The unit half-disc is a compact H-hull with mapping-out function $z+z^{-1}$, so its half-plane capacity is one. The monotonicity of half-plane capacity and its scaling rule now giveThus the assertion holds with the universal constant $c=1$ under this normalization.