= Solution
Let $f=g_K^{-1}$. The reflection at infinity used in part (a) gives analytic expansions there for both inverse maps. In particular,
$$
f(w)=w-\frac{a_K}{w}+O(w^{-2}),
$$
uniformly for large $w\in\mathbb H$, including approach close to the real axis.
Fix $R<\infty$. If $g_K$ were unbounded on $D\cap\{|z|\leq R\}$, there would be points $z_n$ there with $w_n=g_K(z_n)$ tending to infinity in modulus. The inverse expansion would imply $z_n=f(w_n)=w_n+O(1/w_n)$, contradicting boundedness of $z_n$. Thus \b[$g_K$ is bounded on every bounded portion of its domain], including points arbitrarily near a rough hull boundary. This is the <inverse-at-infinity criterion for local boundedness of a mapping-out function>.
On the region $|z|>R$ for sufficiently large $R$, the expansion of $g_K$ gives $g_K(z)-z=O(1/z)$. On the remaining bounded region, the preceding bound for $g_K$ and the bound for $z$ give a finite bound for their difference. Therefore
$$
\boxed{\sup_{z\in D}|g_K(z)-z|<\infty.}
$$
The argument uses reflection near infinity only. It does not assume that the real part of the map extends continuously at every point of an arbitrary hull; the <sharp displacement bound for a compact H-hull> is a further quantitative version of this boundedness.
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