Gromov–Hausdorff distance between a single point and a nonempty compact subset K of a metric space is equal to half the diameter of K I need to show that the Gromov–Hausdorff distance between a single point and a nonempty compact subset $K$ of a metric space is equal to half the diameter of $K$, and I don't see how to do this.
In this paper: https://arxiv.org/pdf/2110.06101.pdf(p. 2), the property is mentioned without a proof as if it were standard but I can't seem to find any reference for it.
Any help would be much appreciated.

As I said in my comment, you have to prove two inequalities:
$$
d_{GH}(K, \{p\})\le D/2,
$$
$$
d_{GH}(K, \{p\})\ge D/2,
$$
where $D=diam(K)$. I will prove the former and leave it to you to establish the latter (it is the easier part).
Given a compact metric space $(K,d_K)$, I define a new metric space $(M,d_M)$ as
$$
M:= K\sqcup \{m\}, d_M(k,m)=D/2 \quad \forall k\in K,
$$
$$
d_M(x,y)= d_K(x,y), \forall x, y\in K. 
$$
Let's check that $(M,d_M)$ is indeed a metric space. The only non-obvious part is to check the triangle inequalities for triples
$x, y, z$, where $z=m$ and $x, y\in K$:
$$
d_M(x,y)\le D= d(x,z)+d(y,z),
$$
$$
d_M(x,z)=D/2\le d_M(x,y) + d_M(y,z)= d_M(x,y) + (D/2). 
$$
So, $(M,d_M)$ is a metric space and the inclusion map $(K,d_K)\to (M,d_M)$ is an isometric embedding.
Now, define a map
$$
f: K\sqcup \{p\}\to M, 
$$
where $f|_K$ is the identity embedding $K\to K$ and $f(p)=m$. By the construction, $f$ restricts to isometric embeddings on both
$K$ and $\{p\}$ and the Hausdorff distance between the images of $K$ and $\{p\}$ is exactly $D/2$. Thus,
$$
d_{GH}(K, \{p\})\le D/2.
$$
