Let $(E, d)$ be a metric space. Then $E$ is precompact if and only if $$(\forall\varepsilon>0)(\exists F\subset E\wedge|F|\in\mathbb{N})(\forall x\in E)(d(x,F)<\varepsilon).$$
Dieudonne's proof goes as follows:
Suppose we have an open covering $(U_\lambda)_{\lambda\in L}$ of $E$ such that no finite subfamily is a covering of $E$. We define by induction a sequence $(B_n)$ of balls in the following way: from assumption, it follows that the diameter of $E$ is finite, and by multiplying the distance on $E$ by a constant, we may assume that $\delta(E)<1/2$, hence $E$ is a ball $B_0$ of radius $1$...
I don't understand how $B_0$ is being defined here. Precompactness implies that, for every $\varepsilon>0$, there exists a finite family $(y_j)_{1\leq j\leq m}$ of elements of $E$ such that $$(\forall x)\left(x\in E\Rightarrow\inf_{1\leq j\leq m}d(x,y_j)<\varepsilon\right).$$ Why does this imply that $\delta(E)<1/2$ and how does this show that $E=B_0$ where $B_0$ is a ball with radius 1? What is the center of $B_0$? (how can you say something is a ball without specifying a radius and a center?)
Can someone please help define $B_0$ explicity?
It does not imply that $\delta(E)<\frac12$. But precompactness implies that $\delta(E)<\infty$. And the Dieudonné multiplies the original distance $d$ defined on $E$ by a constant $K>0$, thereby getting a new distance $d_K$. And $K$ is chosen so that the diameter of $E$ is smaller than $\frac12$. Then $B_0$ is any ball with radius $1$ (with respect to $d_K$).