Prove the inequality concerning diameter in $\mathbb{R}^2$ A is a closed and bounded subset of $\mathbb{R}^2$. Prove that $\sqrt{2}r_0 \le \operatorname{diam}(A) \le 2r_0$, where $r_0 = inf\{r | \exists x\in \mathbb{R}^2 : A \subseteq D_r(x) \}$, where $D_r(x) = \{y \in R \mid d(x, y)\le r\}$. I have problem to prove that $\sqrt{2}r_0 \le \operatorname{diam}(A)$. 
As for the $\operatorname{diam}(A) \le 2r_0$ I think that if $\operatorname{diam}(A) = d(x_0, y_0)$, then $d(x_0, y_0) \le d(x_0, a) + d(a, y_0) <=2r_0$, where $a$ is the center of $D_{r_0}(a)$ such that $A \subseteq D_{r_0}(a)$
 A: This is a bit laborious and I would like to see a simpler approach, but I think it works. In particular, I'm not sure we need the lemma, but it is convenient for visualizing things.
We want to show that $\operatorname{diam}(A) \ge \sqrt{2} r_0$.
First, we can assume $|A| \ge 2$, which means $r_0 > 0$; if not, the result is trivial.
Lemma. There's a minimal bounding disk for $A$ which touches $A$ on the edge of the disk.
That is, there are $x_0 \in \Bbb{R}^2$ and $a_0 \in A$ such that $A \subseteq D_{r_0}(x_0)$ and $a_0 \in \partial D_{r_0}(x_0)$.
Proof. 
Take $B$ to be a closed rectangular region which contains $A$ (since $A$ is bounded).
For $x \in \Bbb{R}^2$, define $r(x) = \operatorname{inf} \{ r | A \subseteq D_r(x) \}$, so that
$r_0 = \operatorname{inf} \{ r(x) | x \in \Bbb{R}^2 \}$.
I claim that in fact we only need to look at $x \in B$: 
$r_0 = \operatorname{inf} \{ r(x) | x \in B \}$.
This is true because for any $x$ which is not in $B$, we can take the nearest point $x' \in B$ to $x$,
and $x'$ is an improvement on $x$: $d(x', a) < d(x, a)$ for all $a \in A$, so if $A \subseteq D_r(x)$ then we also have $A \subseteq D_r(x')$.
Also, $r$ is a continuous function $\Bbb{R}^2 \to [0, \infty)$, because if we move $x$ slightly, then all the distances $d(x, a)$ change only slightly.
Since $B$ is compact, this means there is some $x_0 \in B$ such that $r_0 = r(x_0)$.
Also, if $A$ is disjoint from $\partial D_{r_0}(x_0)$, then since $A$ is compact, there must be a positive distance $\epsilon$ between $A$ and $\partial D_{r_0}(x_0)$.
But then we can shrink $r_0$ by $\epsilon$, which contradicts the definition of $r_0$.
So there must be some point $a_0 \in A \cap \partial D_{r_0}(x_0)$.
This finishes the lemma.
For convenience, we can shift and rotate $A$ so that $x_0 = (0, 0)$ and $a_0 = (r_0, 0)$.
Now I claim that for some $a \in A$, $d(a_0, a) \ge \sqrt{2} r_0$, which will finish the proof.
We'll show that if there is no such $a$, then $A$ lies inside a smaller circle, which again contradicts the definition of $r_0$.
So suppose there is no such $a$.
Taking $Z = D_{\sqrt{2} r_0}(a_0)$, this means $A$ lies inside the interior $\operatorname{Int} Z$.
Since $A$ does not meet the closed exterior $(\operatorname{Int} Z)^c$ and $A$ is compact, there is a positive distance $\epsilon$ between those two sets.
That is, taking $Z'$ to be the smaller disk $D_{\sqrt{2} r_0 - \epsilon}(a_0)$, we have $A \subseteq Z'$.
Note that $0 < \epsilon < \sqrt{2} r_0$.
Now $A$ lies inside both $D_{r_0}(x_0)$ and $Z'$, so we look at the intersection $W = D_{r_0}(x_0) \cap Z'$.
The intersection of $\partial Z$ with $\partial D_{r_0}(x_0)$ is $\{ (0, r_0), (0, -r_0) \}$.
(This is why we picked $\sqrt{2} r_0$.
Now is a good time to draw a picture.)
So the intersection of $\partial Z'$ with $\partial D_{r_0}(x_0)$ is $\{ (\delta, t), (\delta, -t) \}$, where $0 < \delta < r_0$ and $0 < t < r_0$.
$W$ is a slice of $Z'$ on the left and a slice of $D_{r_0}(x_0)$ on the right, joined by the line segment from $(\delta, t)$ to $(\delta, -t)$.
Both the slices are smaller than half-disks, which means $W$ lies inside the disk $D_t(\delta, 0)$ and therefore $A \subseteq D_t(\delta, 0)$.
But $t < r_0$, which is a contradiction.
This completes the proof.
