Let $B_r(y)\subset\mathbb C$ denote the open disk of radius $r$ and center $y\in\mathbb C$, and  $B_r:=B_r(0)$. Let $h:B_1\to U$ a homeomorphism (e.g. a Riemann mapping) . Then  $\Gamma_r:=h(\partial B_r)$ is a simple closed curve in $U$ for $0<r<1.$  Moreover
 
 * For any $\epsilon>0$ there exists $0<\rho<1$ such that for all $\rho<r<1$ there holds $\Gamma_r \subset \partial U  + B_\epsilon$. Indeed
$h^{-1}\big(U\setminus(\partial U + B_\epsilon)\big)$ is a compact
   subset of $B_1$, so it is included in some $B_\rho$ for $\rho<1$,
   whence  $U\setminus(\partial U + B_\epsilon)\subset h(B_\rho)$, that
   is $\Gamma_r\subset h(B_1\setminus B_\rho)\subset \partial U +
   B_\epsilon$ for all $\rho<r<1$.
 * For any $\epsilon>0$ there exist $\sigma>0$ such that for all $\sigma<r<1$ there holds $\partial U\subset \Gamma_r + B_\epsilon$. This follows easily by compactness, arguing by contradiction: otherwise, there would exist  $\epsilon_0>0$ such that for a sequence $r_k <1$, $r_k\to 1$ and $y_k\in\partial U$ one has $\emptyset\neq \partial U\setminus (\Gamma_{r_k} + B_{\epsilon_0})\ni y_k$, that is $ B_{\epsilon_0}(y_k)\cap \Gamma_{r_k}=\emptyset.$ Up to a sequence , $y_k\to y_\infty\in\partial U$ and $ B_{\epsilon_0/2}(y_\infty)\cap \Gamma_{r_k}=\emptyset$ eventually, a contradiction because since $y\infty\in\partial U$, any nbd of $y_\infty$ contains points of $\gamma_r$ for all $r<1$ close enough to $1$.

Therefore the Hausdorff distance is $d_{\mathcal H}(\partial D, f(\partial B_r))=o(1)$ as $r\to 1$.