I'm currently reading the proof that brownian motion exists in Peter Mörters and Yuval peres's book Brownian Motion. In their construction they use an inductive step in the beginning in which I don't understand a step. For completness I've put all of the proof up to the unclear step in the question. For the exact definition of brownian motion used in the book see my question here.
We first construct Brownian motion on the interval [0,1] as a random element on the space $\mathbf{C}[0,1]$ of continuous functions on $[0,1] .$ The idea is to construct the right joint distribution of Brownian motion step by step on the finite sets $$ \mathcal{D}_{n}=\left\{\frac{k}{2^{n}}: 0 \leqslant k \leqslant 2^{n}\right\} $$ of dyadic points. We then interpolate the values on $\mathcal{D}_{n}$ linearly and check that the uniform limit of these continuous functions exists and is a Brownian motion.
To do this let $\mathcal{D}=\bigcup_{n=0}^{\infty} \mathcal{D}_{n}$ and let $(\Omega, \mathcal{A}, \mathbb{P})$ be a probability space on which a collection $\left\{Z_{t}: t \in \mathcal{D}\right\}$ of independent, standard normally distributed random variables can be defined. Let $B(0):=0$ and $B(1):=Z_{1} .$ For each $n \in \mathbb{N}$ we define the random variables $B(d), d \in \mathcal{D}_{n}$ such that
$\hspace{1cm}(1)$ for all $r<s<t$ in $\mathcal{D}_{n}$ the random variable $B(t)-B(s)$ is normally distributed with mean zero and variance $t-s,$ and is independent of $B(s)-B(r)$
$\hspace{1cm}(2)$ the vectors $(B(d): d \in \mathcal{D}_{n})$ and $(Z_{t}: t \in \mathcal{D} \backslash \mathcal{D}_{n})$ are independent.
Note that we have already done this for $\mathcal{D}_{0}=\{0,1\} .$ Proceeding inductively we may assume that we have succeeded in doing it for some $n-1 .$ We then define $B(d)$ for $d \in \mathcal{D}_{n} \backslash \mathcal{D}_{n-1}$ by $$ B(d)=\frac{B\left(d-2^{-n}\right)+B\left(d+2^{-n}\right)}{2}+\frac{Z_{d}}{2^{(n+1) / 2}} $$ Note that the first summand is the linear interpolation of the values of $B$ at the neighbouring points of $d$ in $\mathcal{D}_{n-1} .$ Therefore $B(d)$ is independent of $\left(Z_{t}: t \in \mathcal{D} \backslash \mathcal{D}_{n}\right)$ and the second property is fulfilled.
Now my question is: Why is $B(d)$ independent of $\left(Z_{t}: t \in \mathcal{D} \backslash \mathcal{D}_{n}\right)$
We can prove by induction on $n$ that if $d\in\mathcal D_n$, then $B(d)$ is a function of $\left(Z_d,d\in\mathcal D_n\right)$. It is true for $n=1$, an if it is true for $n-1$ then take $d\in\mathcal D_n\setminus \mathcal D_{n-1}$ and using $$ B(d)=\frac{B\left(d-2^{-n}\right)+B\left(d+2^{-n}\right)}{2}+\frac{Z_{d}}{2^{(n+1) / 2}}, $$ we notice that $d-2^{-n}$ and $d+2^{-n}$ belong to $\mathcal D_{n-1}$ (since $d\notin \mathcal D_{n-1}$, $d=(2k-1)2^{-n}$) and by the induction assumption, $B\left(d-2^{-n}\right)$ and $B\left(d+2^{-n}\right)$ are functions of $\left(Z_d,d\in\mathcal D_{n-1}\right)$ hence of $\left(Z_d,d\in\mathcal D_n\right)$.