The Hausdorff distance between the closed unit disk $D^2$ of $\mathbb R^2$ (equipped with the standard Euclidean distance) and its boundary circle $S^1$ is obviously one.

Interestingly, the Gromov-Hausdorff distance between $D^2$ and $S^1$ is smaller. I claim that it is $\sqrt 3/2$. Below are partial results to support this.

First some notation. A *correspondence* between two metric spaces $X$ and $Y$ is a subset $R \subset X \times Y$ such that the natural projections of $R$ onto $X$ and $Y$ are both surjective. The *distortion* of $R$ then is
$$
\operatorname{dis}(R) := \sup\{|d_X(x,x')-d_Y(y,y')|\, : \, (x,y), (x',y') \in R\} \ .
$$
It can be shown that the Gromov-Hausdorff distance between $X$ and $Y$ is equal to $\frac{1}{2}\inf\operatorname{dis}(R)$, where the infimum is taken over all correspondences between $X$ and $Y$.

Here is a correspondence $R$ between $D^2$ and $S^1$ with distortion $\sqrt 3$. In $R$ are all pairs $(x,x)$ if $x \in S^1$ and also all pairs $(x,c)$ if $c$ is the "center" of the sector to which $x$ belongs as indicated in the following figure ($\sqrt 3$ is the distance between two different black center points):

Below is a sketch of a proof that $\operatorname{dis}(R) \geq \sqrt 3$ in case $R$ is a correspondence between $D^2$ and $S^1$ for which $(x,x) \in R$ for all $x \in S^1$: Color $D^2$ with three colors. First color three arcs of length $2\pi/3$ on $S^1$ with different colors as in the figure above. Now for any $x \in D^2\setminus S^1$ assign to it the color of $y \in S^1$ if $(x,y) \in R$ (pick one in case there are multiple choices). As an application of Sperner's lemma there are three arbitrarily close points of different color. So these three points correspond to points on different arcs of the boundary. This forces $\operatorname{dis}(R) \geq \sqrt 3$. (The assumption $(x,x) \in R$ for $x \in S^1$ is used for the boundary condition in Sperner's lemma.)

Does someone have an idea how to get rid of the condition $(x,x) \in R$ for $x \in S^1$ in the statement above? This would then prove that the Gromov-Hausdorff distance between $D^2$ and $S^1$ is indeed $\sqrt 3/2$.