We have $D_n = \left\{\sum_{j=1}^{n} \frac{x_j}{2^j}, (x_j)_{j \in \llbracket 1,n \rrbracket} \in \{0,1\}^n\right\}$ and $\pi_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n}$.
Let $n \in \mathbb{N}^{\star}$. Justify $x \in D_n \Longleftrightarrow 2^n x \in \llbracket 0, 2^n - 1 \rrbracket$.