How can I show that the "binary digit maps" $b_i : [0,1) \to \{0,1\}$ are i.i.d. Bernoulli random variables?
Let $N\in\mathbb{N}$ be arbitrary. It suffices to show that $\left(b_{1},\dots,b_{N}\right)$ is an $N$-tuple of independent Bernoulli variables. For each $n\in\left\{ 1,\dots,N\right\} $, we have $$ \left[0,1\right)=\biguplus_{\ell=1}^{2^{n}}\left[\left(\ell-1\right)\cdot2^{-n},\ell\cdot2^{-n}\right) $$ and the random variable $b_{n}$ is constant on each of the intervals $$ \left[\left(\ell-1\right)\cdot2^{-n},\ell\cdot2^{-n}\right)=\biguplus_{j=\left(\ell-1\right)\cdot2^{N-n}}^{\ell\cdot2^{N-n}}\left[\left(j-1\right)\cdot2^{-N},j\cdot2^{-N}\right). $$ More precisely, we have $$ b_{n}\equiv\left(\left(\ell-1\right)\,{\rm mod}\,2\right)\qquad\text{ on }\left[\left(\ell-1\right)\cdot2^{-n},\ell\cdot2^{-n}\right)\qquad\left(\dagger\right) $$ and hence $$ b_{n}\equiv\left(\left(\ell-1\right)\,\mod\,2\right)\qquad\text{ on }\left[\left(j-1\right)\cdot2^{-N},j\cdot2^{-N}\right)\qquad\text{ if }\left(\ell-1\right)\cdot2^{N-n}\leq j\leq\ell\cdot2^{N-n}. $$
Observe the for each $j\in\left\{ 1,\dots,2^{N}\right\} $, there is some $\ell\in\left\{ 1,\dots,2^{n}\right\} $ such that $\left(\ell-1\right)\cdot2^{N-n}\leq j\leq\ell\cdot2^{N-n}$. This easily implies that the map $$ f:\left(b_{i},\dots,b_{N}\right):\left[0,1\right)\to\left\{ 0,1\right\} ^{N} $$ is constant on each of the $2^{N}$ intervals $\left[\left(j-1\right)\cdot2^{-N},j\cdot2^{-N}\right)$ for $j\in\left\{ 1,\dots,2^{N}\right\} $.
We finally observe that $f$ is surjective, because for $k:=\left(k_{1},\dots,k_{N}\right)\in\left\{ 0,1\right\} ^{N}$ arbitrary, we can set $$ x:=\sum_{\ell=1}^{N}k_{\ell}2^{-\ell}+\sum_{\ell=N+1000}^{\infty}2^{-\ell}\in\left[0,1\right). $$ Then the right hand side of the above equation directly yields the infinite binary expansion of $x$ and hence $b_{n}\left(x\right)=k_{n}$ for all $n\in\left\{ 1,\dots,N\right\} $, which means $f\left(x\right)=k$.
As the cardinality $\left|\left\{ 0,1\right\} ^{N}\right|=2^{N}$ is equal to the cardinality $$ 2^{N}=\left|\left\{ 1,\dots,2^{N}\right\} \right|=\left|\left\{ \left[\left(j-1\right)\cdot2^{-N},j\cdot2^{-N}\right)\,\mid\, j\in\left\{ 1,\dots,2^{N}\right\} \right\} \right|, $$ this implies that for each $k=\left(k_{1},\dots,k_{N}\right)\in\left\{ 0,1\right\} ^{N}$, there is a unique $j_{k}\in\left\{ 1,\dots,2^{N}\right\} $ such that $f\equiv k$ on $\left[\left(j_{k}-1\right)\cdot2^{-N},j_{k}\cdot2^{-N}\right)$. But this ensures \begin{eqnarray*} \lambda\left(\left\{ x\in\left[0,1\right)\,\mid\, b_{1}\left(x\right)=k_{1},\dots,b_{N}\left(x\right)=k_{N}\right\} \right) & = & \lambda\left(\left[\left(j_{k}-1\right)\cdot2^{-N},j_{k}\cdot2^{-N}\right)\right)\\ & = & 2^{-N}\\ & = & \prod_{j=1}^{N}\lambda\left(\left\{ x\in\left[0,1\right)\,\mid\, b_{j}\left(x\right)=k_{j}\right\} \right), \qquad (1) \end{eqnarray*} where the last equality is due to $\left(\dagger\right)$, which implies $$ \left\{ x\in\left[0,1\right)\,\mid\, b_{j}\left(x\right)=0\right\} =\biguplus_{\ell=0}^{2^{j-1}-1}\left[2\ell\cdot2^{-j},\left(2\ell+1\right)\cdot2^{-j}\right) $$ and hence $$ \lambda\left(\left\{ x\in\left[0,1\right)\,\mid\, b_{j}\left(x\right)=0\right\} \right)=\sum_{\ell=0}^{2^{j-1}-1}\lambda\left(\left[2\ell\cdot2^{-j},\left(2\ell+1\right)\cdot2^{-j}\right)\right)=2^{-j}\cdot2^{j-1}=\frac{1}{2} \qquad (2) $$ as well as $$ \lambda\left(\left\{ x\in\left[0,1\right)\,\mid\, b_{j}\left(x\right)=1\right\} \right)=1-\lambda\left(\left\{ x\in\left[0,1\right)\,\mid\, b_{j}\left(x\right)=0\right\} \right)=\frac{1}{2}. \qquad (3) $$ In summary, equations (1)-(3) establish the claim.