Lemma 1

Suppose the metric space X does not satisfy the first axiom of countability. Then there exists a real number $\varepsilon > 0$ and an uncountable subset A in X such that $d(x, y) \geq \varepsilon$ for all distinct $x, y$ in A.

For every integer $n \geq 1$, let $\mathfrak{F}_n$ be the set (ordered by inclusion) of subsets D of X such that $d(x, y) \geq \frac{1}{n}$ for distinct $x, y$ in D. The set $\mathfrak{F}_n$ is of finite character, hence possesses a maximal element $D_n$ (S, III, § 4, No. 5). Then for all $y \in X$, there exists a point $x$ in $D_n$ such that $d(x, y) < \frac{1}{n}$, by virtue of the maximal character of $D_n$.

Put $D = \bigcup_n D_n$; the set D is then dense in X, and since X does not satisfy the first axiom of countability, D is not countable, and so one of the $D_n$ is not countable.

Q.E.D.

By lemma 1, applied to $B_0$, there exists an uncountable subset $A_0$ of $F_0$ and a number $\varepsilon > 0$ such that $\|x\| \leq 1$ and $\|x - y\| \geq \varepsilon$ for distinct $x, y$ in $A_0$. We have $\mathrm{pr}_0(E) = F_0$ and hence there exists a subset A in E such that $\mathrm{pr}_0$ induces a bijection from A onto $A_0$.