= 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 $z_n\equiv D\alpha\pmod{q^n}$ and $z_n\equiv D\beta\pmod{r^n}$. Put $x_n=z_n/D$. Then
$$
|x_n-\alpha|_q\le|D|_q^{-1}q^{-n}\to0,\qquad
|x_n-\beta|_r\le|D|_r^{-1}r^{-n}\to0.
$$
This 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:
$$
\boxed{y_n=\frac{n!}{1+(n!)^2}.}
$$
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 $|y_n|_p\to0$ for each prime as well. The assertion concerns each fixed prime separately, rather than <uniform convergence> over all primes.
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