Proof that 3 is false.
List the elements of the union of finite cartesian products of {0,1} by increasing $n$ in layers.
Within layer list them by increasing size of the corresponding binary number:
0, 1 $(n=1)$
00, 01, 10, 11 $(n=2)$
000, 001, 010, 011, 100, 101, 110, 111 $(n=3)$
0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 $(n=4)$
....
Now label the list using the natural numbers, first by layer, then moving to the next layer.
The elements in the $n$-th layer will be labeled by $2$$n$-$1$ through $2$$n+1$-$2$.
This provides a bijection between N and all elements of the union of finite cartesian products of {0,1}, so the union of finite cartesian products of {0,1} is countable.
And I think it is done without the assistance of my favourite set theoretical bogeyman, Axiom of Choice.
P.S.
Regarding point 1 in the original question, is there any relation between the infinite countable product of {0,1} and w1 (the first uncountable ordinal)?