= Solution
Write $g=g_K$ and $f=g^{-1}$. We use the <real boundary bounds for a unit-disc H-hull> from part (c), including their reflected version. Set
$$
\alpha=\lim_{x\uparrow-1}g(x),\qquad
\beta=\lim_{x\downarrow1}g(x).
$$
Strict monotonicity and those bounds give $-2\leq\alpha\leq-1$ and $1\leq\beta\leq2$. The inverse $f$ extends across $\mathbb R\setminus[\alpha,\beta]$ and maps these intervals onto $(-\infty,-1)$ and $(1,\infty)$.
Consider the <holomorphic function> $F(w)=w-f(w)$ on $\mathbb H$. If $w$ approaches a real $u$ outside $[\alpha,\beta]$, its inverse tends to a real $x$ with $|x|>1$. Part (c) gives
$$
|u-f(u)|=|g(x)-x|\leq\frac1{|x|}\leq1.
$$
If $u\in[\alpha,\beta]$, every cluster point of $f(w)$ as $w\to u$ lies in the closed unit disc. To justify this without a boundary regularity assumption, first note that $f(w)$ cannot tend to infinity for bounded $w$, since $g(z)=z+O(1/z)$ there. An interior cluster point in $H$ would be mapped to the real number $u$, impossible for $g:H\to\mathbb H$. A real cluster point $x$ with $|x|>1$ would, by the reflected extension and strict monotonicity, have $g(x)$ outside $[\alpha,\beta]$, also impossible. All remaining finite boundary points lie in $K$ or in $[-1,1]$, hence in the unit disc.
It follows for every such $u$ that
$$
\limsup_{\substack{w\to u\\w\in\mathbb H}}|F(w)|
\leq |u|+1\leq3.
$$
Finally $f(w)=w+O(1/w)$ at infinity, so $F(w)\to0$ there. Apply the <maximum modulus principle> on large upper half-discs, using the boundary limsup just obtained, to conclude $|F(w)|\leq3$ everywhere in $\mathbb H$. Taking $w=g(z)$ proves \b[the <sharp displacement bound for a compact H-hull>, including irregular hulls]:
$$
\boxed{|g_K(z)-z|\leq3\qquad(z\in H).}
$$
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