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There is a universal constant $C$ such that every compact H-hull satisfiesTranslate and scale so the hull lies in a unit half-disc, then use the Brownian representation of half-plane capacity and the $O(y^{-1})$ harmonic measure of that half-disc as viewed from $iy$.

WriteThis is the harmonic measure of $A$ viewed from $z$, and is a bounded harmonic function outside the closed target disc. The disc is contained in $B(0,r_2)$. In the half-plane cut out by a line $L$ through the origin, the probability of reaching $B(0,r_2)$ before $L$, from a point of modulus $R$, is $O(r_2/R)$ uniformly in the direction. One sees this by mapping the half-plane outside the disc with $z\mapsto \log (z/r_2)$ to a half-strip and solving the corresponding Dirichlet problem by a sine series.

Part (b) consequently shows that the angular oscillation of $H_A$ on a large circle tends to zero, uniformly in the Borel set $A$. The same half-strip estimate in the annulus $R-|a|\le |z|\le R+|a|$ shows that shifting the center of that circle by the fixed vector $a$ changes the average of $H_A$ by $o(1)$, uniformly in $A$. Hence, for every $\epsilon >0$, all sufficiently large $R$ satisfyfor every Borel subset $A$ of the target disc.

For $r\ge 1$, let $R_r=[-r,r]\times (0,1]$. The Brownian representation of half-plane capacity givesOn hitting $R_r$ the imaginary part is at most one, while the harmonic measure estimate supplied in the question shows that the probability of reaching a disc of radius $O(r)$ containing $R_r$ is $O(r/y)$. Hence $\mathrm{hcap}(R_r)\le Cr$, the half-plane capacity of a low rectangle bound.

SetThe scaling and translation of half-plane capacity giveswhereas $\mathrm{diam}(A_n)=\sqrt{4+n^{-2}}\to 2$. This supplies the required sequence.