A non-rigorous proof would be the following. Suppose
$$x=x_0,\overline{x_1x_2x_3\dots x_n}$$
Then
$$10^nx=x=x_0x_1x_2x_3\dots x_n,\overline{x_1x_2x_3\dots x_n}$$
so $$10^nx-x=x_0x_1x_2x_3\dots x_n-x_0$$
and
$$x=\frac{x_0x_1x_2x_3\dots x_n-x_0}{10^n-1}$$
where $x_n\in\{0,1,\dots,9\}$
Simple example:
$$x=1,234234234\dots$$
then
$$10^3 x=1234,234234\dots$$
so
$$(10^3-1)x=1234-1$$
$$x=\frac{1233}{999}$$
If you want to get more rigorous, you can use the series expansion of a number, but, all in all, the proof's essence won't differ much.
ADD In the more general case
$$x=x_0,y_1y_2y_3\dots y_n\overline{x_1x_2x_3\dots x_n}$$
note
$$x=x_0,0\dots 0\overline{x_1x_2x_3\dots x_n}+0,y_1y_2y_3\dots y_n$$ and consider
$$x'=x_0,0\dots 0\overline{x_1x_2x_3\dots x_n}$$ The shifting is then of $10^{m+n}$, and we obtain the sum of two rational numbers, which is rational.