I prove it by contradiction. Let $A \times B$ is countable. It means we can list down the all the ordered pairs of $A \times B$. So if ordered pairs of the form $(a,b)$ are countable (where $a \in A$ and $b \in B$ ), it implies $A$ and $B$ are countable. But here lies a contradiction since $A$ is uncountable. So our supposition is wrong and $A \times B$ is uncountable.
Is my reasoning alright? What can be a better way to prove this ?
Your argument isn’t really complete: you say that countability of $A\times B$ implies countability of $A$ and of $B$, but you don’t actually justify this.
In any case you’re working much too hard. Fix $b\in B$, and define
$$f:A\to A\times B:a\mapsto\langle a,b\rangle\;;$$
it’s very easy to verify that $f$ is injective and hence that $|A|\le|A\times B|$. (Note that this argument does require that $B$ be non-empty; this is crucial, since $A\times\varnothing=\varnothing$ is actually countable.)