= Solution
\b[A nearly closed semicircular slit makes the constant sharp.] Put
$$
m(z)=\frac{z+1}{1-z},\qquad
K_L=m^{-1}(\{iy:0<y\leq L\}),\qquad L>0.
$$
This is the unit-circle arc attached at $-1$ and ending just short of $1$. Its complement is a <simply connected domain>. There remains a narrow passage near $1$ connecting the interior bay to infinity. Thus $K_L$ is a <compact H-hull> inside the closed unit disc.
For an explicit check, let $S=\sqrt{1+L^2}$ and let $q_L(z)=\sqrt{m(z)^2+L^2}$ denote the <mapping-out function of a vertical slit> applied to $m(z)$, with the branch asymptotic to its argument in the slit half-plane. Its normalized map is
$$
g_{K_L}(z)=1+\frac{L^2}{S^2}-\frac{2}{S\,[q_L(z)+S]}.
$$
The final <Möbius transformation> takes the image of the original infinity to infinity and gives <hydrodynamic normalization at infinity>. For a fixed $z$ in the open unit half-disc, $\operatorname{Re}m(z)>0$, so $q_L(z)\sim L$ and $g_{K_L}(z)\to2$ as $L\to\infty$. Now take $z_\delta=-1+\delta+i\delta$, with $0<\delta<1$. First let $L\to\infty$, then let $\delta\downarrow0$. The <nearly closed semicircular slit> gives
$$
\boxed{\lim_{\delta\downarrow0}\lim_{L\to\infty}
|g_{K_L}(z_\delta)-z_\delta|=3.}
$$
Every constant smaller than three therefore fails for some member of this family and some interior point. The bound need not be attained at an interior point of a single fixed hull.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-203-nearly-closed-hull.png]
{title=A nearly closed semicircular slit and a marked point inside its bay}
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