Let $d=\operatorname{diam}(A)$. 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 $A\subseteq\{z\in\mathbb H:|z-x|\leq d\}$.

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 give
$$
\operatorname{hcap}(A)\leq\operatorname{hcap}\{z\in\mathbb H:|z-x|\leq d\}=d^2.
$$
Thus the assertion holds with the universal constant $c=1$ under this normalization.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 1 / b / i / Solution

If compact H-hulls satisfy $A\subseteq C$, then
$$
\operatorname{hcap}(A)\leq\operatorname{hcap}(C).
$$
This follows from the <Brownian representation of half-plane capacity> or from composition of their mapping-out functions.

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 Incoming links: Monotonicity of half-plane capacity