= Solution
The capacity in this question is <harmonic capacity from infinity in the upper half-plane>, which has units of length; it is distinct from <half-plane capacity>, which has units of length squared.
Here is a proof of existence that also works with irregular real attachments. The function
$$
u(z)=\mathbb P_z(B_{T(H)}\in K)
$$
is bounded and <harmonic> on $H$, by the <Strong Markov property> and the mean-value characterization of <harmonic functions>. Therefore $u\circ g_K^{-1}$ is a bounded <harmonic function> on the <complex upper half-plane>, with a <Poisson kernel> representation
$$
u(g_K^{-1}(w))
=\int_{\mathbb R}\frac{\operatorname{Im}w}
{\pi|t-w|^2}\,f(t)\,dt,\qquad 0\le f\le1.
$$
The boundary function $f$ vanishes outside a bounded interval. Indeed, far enough along either real ray the original domain contains a half-disc neighbourhood, and $g_K$ extends there; the probability of hitting the bounded hull before the real boundary tends to zero as the starting point approaches that ray. Applying the same dominated-limit calculation as in part (i) yields
$$
\boxed{\operatorname{cap}(K)=\int_{\mathbb R}f(t)\,dt<\infty.}
$$
When the intrinsic boundary pieces landing on the hull are identified, $f$ is their indicator almost everywhere and this is the length of their image under $g_K$. For ordinary finite slit hulls this is exactly the image of $\delta H\setminus H_0$, since real attachment endpoints have zero <harmonic measure>. The Poisson representation avoids requiring that boundary identification in the general existence argument.
If $K\subset K'$, couple the two exit events using the same <planar Brownian motion>, stopped at its first hit of the real axis. Any path that hits $K$ before the real axis also hits $K'$ before the real axis. Consequently
$$
\mathbb P_{iy}(B_{T(\mathbb H\setminus K)}\in K)
\le
\mathbb P_{iy}(B_{T(\mathbb H\setminus K')}\in K'),
$$
and taking the limits proves \b[monotonicity of this capacity].
For a half-disc of radius $r$ centred at $b\in\mathbb R$, the <mapping-out function> is
$$
g(z)=b+(z-b)+\frac{r^2}{z-b}.
$$
Its semicircular boundary maps onto $[b-2r,b+2r]$, of length $4r$. Hence its <harmonic capacity from infinity in the upper half-plane> is $4r$. With
$$
\operatorname{rad}(K)=\inf\{r>0:K\subset\{z:|z-b|\le r\}
\text{ for some }b\in\mathbb R\},
$$
enclose $K$ in such a half-disc and use monotonicity. Letting the enclosing radius decrease to the infimum proves
$$
\boxed{\operatorname{cap}(K)\le4\operatorname{rad}(K).}
$$
The same conclusion holds if radius is instead measured about a specified real centre.
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