Transfinite cardinals

Please look at the two following 0;1 infinite trees:


The left tree is an infinite set that its members are infinite 0;1 paths, where their complements are included in the tree.

The right tree is an infinite set that its members are infinite 0;1 paths, where their complements are not included in the tree.

Yet by modern mathematics both trees have uncountable distinct paths.

Now define an infinite set that its members are infinite 0;1 paths that are taken from both sides of the left tree, for example:

01101...
11100...
10101...
11001...
00111...
...

where a complement that is not in that set starts with bits 10010...

Such set can be equivalent to the right tree case, which means that defining complements that are not in such set, is not a proof that this set must have countably infinitely many members.

In that case a mapping between members of set N and such considered set, is not necessarily done between countably infinitely many members in both sets.

By following this simple observation, can someone explain why there is a strict distinction between countably infinite and uncountable transfinite cardinals, as done by modern mathematics?

(Infinite sets that their members are infinite 0;1 paths are used here without loss of generality, which means that any n>2 valued infinite sets can be used in this question (but in that case we are not talking about missing complements, but about missing infinite distinct paths)).