Cantor's diagonalisation argument in proving cartesian product of infinitely many countable sets as uncountable There are lot of posts related to argument. But i have some confusion in the proof. My thinking :-
If cartesian product is represented as tuple $\langle s_1, s_2, s_3 .... s_n.....\rangle$
Then say $s_{ij}$ represent jth element of $s_i$ sequence. Then we have
$s_{11}, s_{12}, s_{13}, s_{14}, ...... \\ s_{21}, s_{22}, s_{23}, s_{24}, ...... \\............. \\.........$
now we can cross-out the elements diagonally(like in cantor diagonalisation argument) and say that bijection exists. But this type of argument is used in proving countably infinite union of countable sets is countable. Why the same can not be used in proving for cartesian product of  countably many countable sets? I am not able to perceive the difference. Please clarify

In union, each countable set can be represented as a sequence
In cartesian product can each element be considered a 1 element of product of sequences like 1st tuple as 1 element of S1, S2, S3 .....ie., s11, s21, s31, s41 .. second element as s12, s22, s32, s42 ....
(I am taking A as sequence s1, B as sequence s2, C as sequence s3 .... and not take cartesian product of AXBX... then elements will be product of sequences)
Both arguments can be visualized with an infinite matrix of elements.
For the Cantor argument, view the matrix a countable list of (countably) infinite sequences, then use diagonalization to build a SEQUENCE which does not occur as a row is the matrix. So the countable list of sequences (i.e. rows) is missing a sequence, so you conclude the set of all possible (infinite) sequences is UNCOUNTABLE.
For the countable union of countable sets, each row represents one of the sets in the union, and you do some sort of zig zag enumeration of the matrix entries to show the that the set of ENTRIES in the matrix (i.e. the union of the row sets) is COUNTABLE.
Both matrices have countably many entries, both matrices must be missing possible sequences that can be formed from the elements available. But the union is about the entries in the matrix, while the infinite product is about all possible sequences you could construct.
If each of your factor sets has at least teo elements, you can use the other diagonal argument to construct $t=\langle t_1,t_2,\ldots,t_m,\ldots\rangle$ where $t_m\ne s_{m,m}$ and hence $t\ne s_m$