Will every rational number eventually be in this set?

Let $A_0=\{0\}$. For every $n\ge 0$, let $B_n=\bigcup_{k\ge 0}A_n+k$, and let $A_{n+1}=f(B_n)$, where $f:[0,\infty)\to[0,1):x\mapsto\frac{x}{x+1}$.

Let $q\in\Bbb Q\cap [0,1)$. Can we prove that $q\in A_n$ for some $n\ge 0$? Equivalently, I would like to prove or disprove that: $$\bigcup_{n\ge 0}A_n=Q\cap [0,1)$$

To illustrate: $B_0=\Bbb N$ (including $0$). Then, $A_1=\{0,\frac12,\frac23,\frac34,\ldots\}$. For any natural $k$, we have $\frac12+k=\frac{2k+1}{k}\in B_1$. Thus, $\frac{2k+1}{3k+1}\in A_2$. Et cetera.

Random thought: can this be argued using continued fractions?

Yes, you've constructed the Calkin-Wilf tree. $A_n \mapsto B_n$ corresponds to moving any number of steps to the right. $B_n \mapsto A_{n+1}$ corresponds to moving one step to the left.