In order to give more strength to the induction hypothese let us prove
more generally:
$\exists\alpha\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\alpha_{i}\right)}\right\} _{i\in I}\text{ is independent}\implies\forall\beta\in\left\{ 0,1\right\} ^{I}:\left\{ B_{i}^{\left(\beta_{i}\right)}\right\} _{i\in I}\text{ is independent}$
Assume that the statement is not true. Then some finite subset $J\subseteq I$
can be found such that $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$
for some $\beta\in\left\{ 0,1\right\} ^{I}$ .
Let $J$ be such set and this with minimal cardinality.
Now find a $\beta\in\left\{ 0,1\right\} ^{I}$ such that $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$
and $J_{1}=\left\{ j\in J\mid\alpha_{j}\neq\beta_{j}\right\} $ has
minimal cardinality.
Then $J_{1}\neq\varnothing$. Let $r\in J$ with $\alpha_{r}\neq\beta_{r}$.
$\prod_{j\in J-\left\{ r\right\} }\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)=\mathbb{P}\left(\bigcap_{j\in J-\left\{ r\right\} }B_{j}^{\left(\beta_{j}\right)}\right)=\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)+\mathbb{P}\left(\bigcap_{j\in J-\left\{ r\right\} }B_{j}^{\left(\beta_{j}\right)}\cap B_{r}^{\left(\alpha_{r}\right)}\right)=\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)+\prod_{j\in J-\left\{ r\right\} }\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)\left(1-\mathbb{P}\left(B_{r}^{\left(\beta_{r}\right)}\right)\right)$
contradicting that $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$.
The first equality is a consequence of the minimality of $|J|$ and the third is a consequence of the minimality of $|J_1|$.
We conclude that the statement must be true.
edit to make things clear:
For finite $J\subseteq I$ and $\beta\in\left\{ 0,1\right\} ^{I}$
abbreviate $\mathbb{P}\left(\bigcap_{j\in J}B_{j}^{\left(\beta_{j}\right)}\right)\neq\prod_{j\in J}\mathbb{P}\left(B_{j}^{\left(\beta_{j}\right)}\right)$
by $P\left(J,\beta\right)$.
Assuming that the statement is not true is the same as assuming that
finite sets $J\subseteq I$ exists with $\left\{ \beta\in\left\{ 0,1\right\} ^{I}\mid P\left(J,\beta\right)\right\} \neq\varnothing$.
Choose such $J$ and this with minimal cardinality.
Note that $\beta$ is not fixed yet after this choosing of $J$. That
is the next step to take.
We have $\left\{ \beta\in\left\{ 0,1\right\} ^{I}\mid P\left(J,\beta\right)\right\} \neq\varnothing$
and this enables us to choose $\beta\in\left\{ 0,1\right\} ^{I}$
such that $P\left(J,\beta\right)$ and such that $J_{1}$ has minimal
cardinality.