Show inductively for $k=0,1,\ldots $ that $\tilde B_{k,j}(x):=B_k(x+\frac{k-1}{2}-j), \ \ \ j=0,\ldots,k$
is linearly independent on $x \in [0,1]$.
$B_0(x):=\begin{cases}1, && |x|\le \frac{1}{2}\\0, && |x|>\frac{1}{2}\end{cases}$
$B_{k+1}(x):=\int_{x-\frac{1}{2}}^{x+\frac{1}{2}}B_k(\xi) d\xi, \ \ \ x \in \mathbb R, k=0,1,\ldots$
From this follows that $B_{k+1}'(x)=B_k(x+\frac{1}{2})-B_k(x-\frac{1}{2})$
Base case:
$\tilde B_{0,0}(x)=B_0(x-\frac{1}{2})=1$ for $x \in [0,1]$ because $|x-\frac{1}{2}|\le \frac{1}{2}$.
I need help with the induction step. I want to show that $(\tilde B_{k+1,j})_{j=0,\ldots, k+1}$ is linearly independent, that is if $\sum_{j=0}^{k+1}a_j\tilde B_{k+1,j}=0$ then $0=a_0=a_1=\ldots a_{k+1}$. ($a_0,a_1,\ldots,a_{k+1} \in \mathbb R$)
using the induction hypothesis that $(\tilde B_{k,j})_{j=0,\ldots,k}$ is linearly independent.
I think this helps:
$B_k$ is non-negative and $B_k=0$ for $|x|>\frac{k+1}{2}$
Note first that$$\tilde{B}_{k+1,\,j}^\prime(x)=B_{k+1}^\prime\left(x+\frac{k}{2}-j\right)=B_k\left(x+\frac{k+1}{2}-j\right)-B_k\left(x+\frac{k-1}{2}-j\right)\\=\tilde{B}_{k,\,j-1}(x)-\tilde{B}_{k,\,j}(x),$$provided we define $\tilde{B}_{k,\,k+1}:=0$. If $\sum_{j=0}^{k+1}a_j\tilde{B}_{k+1,\,j}=0$ then, at values of $x$ where derivatives are well-behaved because none of the $B_\ell$ have an argument of $\pm\frac12$,$$0=\sum_{j=1}^{k+1}a_j\tilde{B}_{k+1,\,j}^\prime=\sum_{j=1}^{k+1}a_j(\tilde{B}_{k,\,j-1}-\tilde{B}_{k,\,j})=a_1\tilde{B}_{k,\,0}+\sum_{j=1}^k(a_{j+1}-a_j)\tilde{B}_{k,\,j}.$$By the inductive hypothesis, $a_1=0$ and each $a_{j+1}-a_j=0$, and so the only possible nonzero coefficient is $a_0$. But this must also be $0$, because $\tilde{B}_{k,\,0}$ isn't identically $0$.