@johnny09 This would not be a correct proof.

If you want to proof along this line you have to show that existence of any injection leads to a contradiction. You cannot simply choose the for of $f$.

However, the same argument works for any function, you can fix some $n$ and take $g(m)=f(m,n)$.

In other words, if $\mathbb R\times\mathbb R$ is countable than $$|\mathbb R| \le |\mathbb R\times \mathbb R| \le \aleph_0.$$