 Solution

Choose a nonzero integer $D$ with $D\alpha ,D\beta \in \mathbb{Z}$. By the Chinese remainder theorem, choose $z_n\in \mathbb{Z}$ satisfying $z_n\equiv D\alpha (\mathrm{mod}{q}^n)$ and $z_n\equiv D\beta (\mathrm{mod}{r}^n)$. Put $x_n=z_n/D$. ThenThis proves the requested simultaneous approximation, even when the rational targets have denominators divisible by either prime.

A rational sequence converging to zero at every standard absolute value is:Its ordinary absolute value is at most $1/n!$, hence tends to zero. For a fixed prime $p$ and all sufficiently large $n$, the denominator is congruent to one modulo $p$, so $v_p(y_n)=v_p(n!)\to \infty$. Therefore $|y_n{|}_p\to 0$ for each prime as well. The assertion concerns each fixed prime separately, rather than uniform convergence over all primes.