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...
You suppose that you can choose $s\in [0,1)$ as near to 1 as you want, but it isn't generally true in a $\mathbb{K}$-vector space. Consider for example a $\mathbb{Z}_{p}$-vector space, than $B_{1}(0)=\{0\}$. That's why in general the statement is false.
Anyway, i think your proof works in every $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$-vector space