Any two points inside a circle are within a diameter of each other. In many problems involving the Pigeonhole Principle, we often assume the following lemma:

Lemma: The distance between any two points in a circle of radius $r$ is at most $2r$.

Intuitively, this is obvious: "just stretch out the two points as far as possible until they're diametrically opposite from each other". I was wondering how we could rigorously prove this lemma. I think I can handle proving the $1$-dimensional version as follows:


Easier Lemma: If $x,y \in [-r,r]$, then $|y - x| \leq 2r$.

Proof of Easier Lemma: Without loss of generality, assume that $-r \leq x \leq y \leq r$. Then it suffices to show that $y - x \leq 2r$. Now recall that since $x \geq -r$, we know that $-x \leq r$. But then since $y \leq r$, we can add the previous two inequalities together to obtain $y - x \leq 2r$, as desired. $~~\blacksquare$

When I try to generalize my proof to two dimensions, I run into a snag in the first line, where I try to drop the absolute value with a "without loss of generality". How can I get around this in the $2$-dimensional version?
 A: Thanks to Daniel Fischer's comment, this lemma is actually really trivial to prove. Let's generalize!
Given any metric space $(M, d)$, we define the closed ball of radius $r > 0$ centred at $c \in M$ to be:
$$
B_r[c] = \{x \in M \mid d(x, c) \leq r\}
$$
Using this new notation, we now want prove the following lemma:

Generalized Lemma: If $x,y \in B_r[c]$, then $d(x, y) \leq 2r$.

Proof of Generalized Lemma: Given any $x,y \in B_r[c]$, observe that:
\begin{align*}
d(x,y)
&\leq d(x,c) + d(c,y) &\text{by the triangle inequality} \\
&= d(x,c) + d(y,c) &\text{by symmetry} \\
&\leq r + r &\text{since } x,y \in B_r[c] \\
&= 2r
\end{align*}
as desired. $~~\blacksquare$

Thus, my original lemma can be proved by taking $M = \mathbb R^2$ and defining $d\colon M \times M \to \mathbb R$ by:
$$
d(\vec x, \vec y) = \|\vec x - \vec y \|_2 = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2}
$$
A: Suppose the circle is centered at $(0,0)$
$$x_1^2+y_1^2\leq r^2$$
$$x_2^2+y_2^2\leq r^2$$
$$(x_1-x_2)^2+(y_1-y_2)^2 \leq x_1^2+x_2^2+y_1^2+y_2^2+2|x_1x_2+y_1y_2|$$
$$=x_1^2+x_2^2+y_1^2+y_2^2+2|(x_1,y_1)\cdot(x_2,y_2)|\leq 2r^2+2||(x_1,y_1)||||x_2,y_2||=4r^2$$
