It is easy to see that $ \delta(B_1(0))\le 2$ because $ $ \|x-y\|\le\|x\|+\|-y\|=\|x\|+|-1|\|y\|=\|x\|+\|y\|<1+1=2$ $ so $ \sup\{\|x-y\|:x,y\in B_1(0)\}\le 2$

But to show that $ \delta(B_1(0))\ge 2$ we chose $ a\in B_1(0)\backslash\{0\}$ and $ s\in[0,1)$ and observe that $ \|s\frac{a}{\|a\|}\|=s$ $ \implies s\frac{a}{\|a\|},-s\frac{a}{\|a\|}\in B_1(0)$ and $ \|s\frac{a}{\|a\|}-(-s\frac{a}{\|a\|})\|=2s$

But why does that imply that $ \delta(B_1(0))\ge2s$ ?

This is how I think I understand it: the diameter of $ B_1(0)$ is greater than that of $ B_s(0)$ for all $ s\in[0,1)$ because we it contains it. And the diameter of $ B_s(0)$ is $ 2s$ for each $ s$ , which is true because there exists a sequence $ (\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})$ of pairs of points, with $ 0\le\epsilon_n<s~\forall n$ and $ \lim\limits_{n\rightarrow\infty}\epsilon_n=s$ such that it is always contained in $ B_s(0)$ and $ \lim\limits_{n\rightarrow\infty}d(\epsilon_n\frac{a}{\|a\|},-\epsilon_n\frac{a}{\|a\|})=2s$

I’m not sure if I got it right…