To demonstrate that if $x$ is rational, then $y$ is also rational, we begin by considering the nature of rational numbers. A number $x$ is rational if it can be expressed as a fraction $\frac{p}{q}$, where $p$ and $q$ are integers, and $q \neq 0$. Consequently, the decimal representation of $x$ is either terminating or eventually repeating.

Given the number $x \in (0, 1)$, let its decimal representation be given by:

$$x = 0.a_1a_2a_3\ldots$$

where each $a_i$ is a digit from 0 to 9.

Now, define $y$ as the number whose $n$-th digit after the decimal point is the $2^n$-th digit of $x$. Thus, the decimal representation of $y$ is:

$$y = 0.b_1b_2b_3\ldots$$

where $b_n = a_{2^n}$.

To show that $y$ is rational, it's crucial to show that $y$'s decimal representation is also eventually repeating if $x$'s decimal representation is.

1. **Case 1: $x$ has a terminating decimal**
- If $x$ has a terminating decimal representation, say it ends after $M$ digits, then beyond some point, all digits of $x$ are zero: $a_{k} = 0$ for all $k > M$.
- Consequently, since each $b_n = a_{2^n}$, there exists some $N$ such that for all $n$ with $2^n > M$, $b_n = 0$. Thus, $y$ also has a terminating decimal.

2. **Case 2: $x$ has a repeating decimal**
- If $x$'s decimal representation is eventually repeating with period $T$, there exist $M$ such that the sequence $a_{M+1}, a_{M+2}, \ldots$ repeats every $T$ digits, e.g., $a_{M+k} = a_{M+k+T}$ for all $k \geq 1$.
- Observing $y$, because powers of two increase exponentially, the digits $a_{2^n}$ are eventually past the point $M$, entering the repeating cycle. Therefore, $b_n$ will also enter a repeating cycle since each period for $a_i$ will again align due to repeating properties for large $n$.

The number $y$, constructed from repeating or terminating digits of $x$, inherits these traits, thus $y$ must be eventually repeating or terminating, hence rational.

Therefore, if $x$ is rational, then $y$ is rational. Hence, our proof concludes with the statement:

$$\boxed{\text{If } x \text{ is rational, then } y \text{ is rational.}}$$