 Solution

The implication is false. Take the thin rectanglesEach is a compact H-hull, and $\mathrm{diam}(A_{\epsilon })=\sqrt{4+{\epsilon }^2}\to 2$. We verify that its half-plane capacity nevertheless tends to zero.

Let $K=\{z\in \mathbb{H}:|z|\le 2\}$, which contains all these rectangles. At the Brownian exit time $\tau$ from $\mathbb{H}\setminus A_{\epsilon }$, the exit height is at most $\epsilon$, and a positive exit height requires hitting $K$ before $\mathbb{R}$. ThereforeFor $y>2$, conformal invariance of planar Brownian motion and $g_K(z)=z+4/z$ identify the probability on the right with the harmonic measure of $[-4,4]$ from $i(y-4/y)$ in $\mathbb{H}$. Integrating the Poisson kernel for the upper half-plane givesThe Brownian representation of half-plane capacity now proves the explicit estimateChoose $\epsilon =1/n$ to obtain the required sequence. This is an instance of half-plane capacity of a low rectangle: shrinking height can make half-plane capacity small while horizontal extent stays fixed.