Up to 6 points spaced $r$ apart can fit on a rectangle of dimensions $(r,2r)$ Consider a rectangle of dimensions $(r,2r)$. Is it true that one can place only up to 6 points spaced at least $r$ apart from one another in or on the rectangle?
One can place the 6 points on the four corners and the midpoints of the long sides of the rectangle and it is easily seen that the points are spaced at least $r$ apart from one another.
How would one formally prove that this is impossible with 7 or more points?
 A: Yes, it is true.
Proof : 
Let $ABCD$ be the rectangle where
$$AB=CD=r,\quad AD=BC=2r$$
Let $E,F$ be the midpoint of $AD,BC$ respectively.
In the following, we use the following lemma whose proof is written at the end of the answer : 
Lemma : The only way to place $4$ points spaced at least $r$ apart from one another in or on a square whose side-length is $r$ is to place them on the vertices.
It follows from the lemma that it is impossible to place $5$ or more points spaced at least $r$ apart from one another in or on a square whose side-length is $r$.
Suppose that we can place $7$ points in or on the rectangle.
Let $N(\alpha)$ be the number of points in or on $\alpha$.
Now, let us consider three parts $ABFE$ except the line segment $EF$, the line segment $EF$, and $EFCD$ except the line segment $EF$, and let 
$$\begin{align}a:&=N(ABFE\ \text{except the side $EF$})\\
b:&=N(EF)\\
c:&=N(EFCD\ \text{except the side $EF$})\end{align}$$
Note that $b\le 2$, and we may suppose that $a\ge c$.
If $b=0$, then $a+b+c=7$ and $a\ge c$ imply $a\ge 4$, and $N(ABFE)=4$. It follows from the lemma that $b=2$ which contradicts $b=0$.
If $b=1$, then $a+b+c=7$ and $a\ge c$ imply $a\ge 3$ and $N(ABFE)=4$. It follows from the lemma that $b=2$ which contradicts $b=1$.
If $b=2$, then $a+b+c=7$ and $a\ge c$ imply $a\ge 3$ and $N(ABFE)\ge 5$ which is impossible.$\quad\blacksquare$

Finally, let us prove the lemma : 
Lemma : The only way to place $4$ points spaced at least $r$ apart from one another in or on a square whose side-length is $r$ is to place them on the vertices.
Proof for the lemma : 
Let $Q$ be the square. Since the length of the diagonal is $r\sqrt 2$, the distance of any two of the $4$ points is $\ge r$ and $\le r\sqrt 2$. Take three points $A,B,C$ of the four. They satisfy
$$AB^2+AC^2\ge BC^2\tag1$$
Therefore, $$\angle{BAC}\le 90^\circ\tag2$$
The equality of $(1)(2)$ is attained only when $AB=AC=r,BC=r\sqrt 2$.
Therefore, the four points make a convex quadrilateral $R$. Since every inner angle of $R$ is $\le 90^\circ$, each of the four inner angle is equal to $90^\circ$. These imply that the four points make a square whose side length is $r$.$\quad\square$
