Assume both r and s are rational then $\frac{r}{s}\in\mathbb{Q}$ Currently done with showing for both r+s and r $\times s$ rational.
Now stuck with the following part; If we know that r and s are both rationals what could we say about
$\frac{r}{s}$
Assume both r and s are rationals.
Then we would have by definition r = $\frac{a}{b}$ where a and b are integers and b $\neq$ 0
Similarly;
s = $\frac{c}{d}$ where c and d are integers and d $\neq$ 0
$\frac{r}{s}$ = ($\frac{a}{b}$) /($\frac{c}{d}$)
Hence  $\frac{r}{s}$ = (a $\times d$)/ (b $\times c$)
Since ad and bc are integers, numerator and denumerator are both integers. Integers divided by integers are rational ?
Is this valid proof ?

Yes, it's a valid proof.
The only "error" that may be corrected is that you must assume $s\in\mathbb{Q}\backslash\{0\}$, because if $s=0$ then $r/s$ is undefined and consequently we get that $r/s\notin\mathbb{Q}$.
So, as conclussion for your proof:
$$r\in\mathbb{Q}\phantom{a}, \phantom{a}s\in\mathbb{Q}\backslash\{0\} \Longrightarrow \frac{r}{s}\in\mathbb{Q}.$$
