Prove that, if n sets are countably infinite, then the Cartesian product of all the sets is countably infinite.

Do not cascade indices outwards : x_{1,1} should be preferred to {x_1}_1.

Do not use an alphabetical list to index, $\sigma ((x_{1,A}, x_{2,B}, x_{3,C},...,x_{n,N})) = p_1^A\cdot p_2^B \cdot p_3^C \cdot ... \cdot p_n^N$ is unclear (and implicitely implies $n \leqslant 26$).

Say you want your exponent to be a capital letter, then use (eg) $A_1, A_2,\dots, A_n$. You get the well written definition $$\sigma (x_{1, A_1},\, x_{2,A_2}, x_{3, A_3},\dots,x_{n, A_n})) := p_1^{A_1}\cdot p_2^{A_2} \cdot p_3^{A_3} \cdot \dots \cdot p_n^{A_n}$$.

Your very last part is wrong, which part is...

Clearly, the image is finite. So by definition, the Cartesian product of n sets which are all countably infinite, is itself, countably infinite.

It must be change to this...

Clearly, the image is denumerable. So by definition, the Cartesian product of $n$ sets which are all countably infinite, is itself, countably infinite.

And other part are perfect.