In the Cantor diagonal argument, how does one show that the diagonal actually intersects all the rows in an infinite set? In the Cantor diagonal argument, how does one show that the diagonal actually intersects all the rows in an infinite set? 
Here's what I mean. If we consider any finite sequence of binary representations of length m; constructed in the following manner:


F(n) -> 
 bin(n)
 F(n+2) 
 bin(n+1)

Where bin(n) produces the binary representation of n.
the table isn't square
e.g. m=4.

0000
0010
0100
0110
1000
1010
1100
1110
1111
1101
1011
1001
0111
0101
0011
0001

Clearly not every row meets the diagonal, and so I can flip all the bits of the diagonal; and yes there it is 1111 in the middle of the table.
So if I let the function run to infinity it constructs a similar, but infinite, table with all even integers occurring first (possibly padded out to infinity with zeros if that makes a difference), followed by all odd integers similarly padded: 
Every number in the 'first half' is even, and every number in the 'second half' (if we ever got there) is odd. This table is by construction indexed by the integers.
But, since there are countably infinite even numbers, then naively it seems the table is at least 2 countable infinities long but at most one countable infinity wide, therefore the diagonal is only one countable infinity long and consists of an infinite sequence of zeros; but the table isn't square and so the second half seems to be over an 'event horizon' here and the diagonal never intersects any odd number. 
Hence any argument about flipping bits and it necessarily not being a member of the table seems not to hold. 
What am I missing here?
Thanks
Sean.

By assumption, in the diagonal argument, the rows of your table are indexed by the natural numbers.  In the diagonal argument, you are trying to show that the set of all infinite sequences (indexed by the natural numbers) of $0$s and $1$s is uncountable.  "Countable" means that it is possible to list all of them in a list indexed by the natural numbers.  So to show that your set is uncountable, you suppose that you had a list indexed by the natural numbers, and prove that the list cannot be contain every element of the set.
So it doesn't matter whether there is a "natural" way to create a table that "isn't square", because in the proof, you start by assuming you have a list where the rows and columns are both indexed by the natural numbers.
