I need to prove by induction that if $x, y \in\mathbb{Q}$ with $x < y$, then there is an infinite increasing sequence $\{z_n\}_n$ in $\mathbb{Q}$ such that $x < z_1 < z_2 < \dots < z_n < \dots < y$.
I have tried to use $a_n = x + \frac{y-x}{2^n}$ but am not being able to proceed.
Base case: there is a rational number between the rationals $x$ and $y$. Indeed, $x<z=\frac{x+y}2<y$ and $z$ is a rational.
Induction: if there are $n>0$ rationals between $x$ and $y$, then there are $n+1$ rationals between $x$ and $y$. Indeed,
$$x<z_1<z_2\cdots z_n< y\implies x<\frac{x+z_1}2<z_1<z_2\cdots z_n<y.$$
A non-inductive proof can be "for all $n$, the numbers $\left(1-\frac in\right)x+\frac iny$ with $0<i<n$ are rational, distinct, in $(x,y)$ and increasing".
You are on the right track! However, if you need an increasing sequence take $$z_n = y - \frac{y-x}{2^n}.$$ (note that your sequence stays between $x$ and $y$ but it is decreasing).
Now try to verify that this sequence satisfies the required conditions.
Basic step. Verify $x<z_1<y$.
Inductive step. Verify that for $n\geq 1$, $z_n<z_{n+1}<y$.
If you want a strictly monotonic increasing sequence of rational numbers, then you can also take the $z_{n}=y-\frac {y+z_{n-1}}{2}=\frac {y-z_{n-1}}{2}$ where you can define $z_0=x$.
The basis step and the induction step all hold since sum, difference and division of two non-zero rational numbers always yield a rational number.
