How to count in binary? (basic problem)
I don't really understand the algorithm you're trying to use to count. It seems rather complicated and, instead of trying to correct it, I'm just going to start from scratch.

Really, everything is just like decimal, only there are fewer choices for each digit.  Just like decimal, leading zeroes don't alter the value of the number (0123 is the same number as 123 and 0000123) so we can always add leading zeroes if that helps us.

To add 1 in decimal, we just increase the last digit by&nbsp;1, if it's not already the biggest possible value. If it is the biggest possible value, we replace it with&nbsp;0 and "carry&nbsp;1" to the second-to-last digit. What does that mean? Well, it just means that we increase that digit if we can and, if we can't, we reset it to&nbsp;0 and carry the&nbsp;1 to the previous digit. Eventually, if we started with something like 999, we'll have to say that 999 is the same thing as 0999 when we need to the&nbsp;1 past the left-most&nbsp;9

So, to add&nbsp;1 in binary, we increase the last digit by&nbsp;1, if it's not already the biggest possible value, which is&nbsp;1.  If the last digit is&nbsp;1, we reset it to zero and "carry the&nbsp;1", which we do just like we did before.

This means that it goes as follows:


0
1 (Increase the last digit)
10 (We can't increase the last digit, so we reset it to zero and carry the&nbsp;1. We've run out of digits, so we put a new zero on the left. Then, to carry the&nbsp;1, we add&nbsp;1 to that new digit.)
11 (Increase the last digit)
100 (Reset the last digit to zero, carry the&nbsp;1. Reset the second-to-last digit to zero, carry the&nbsp;1. No digits remain, so put a new zero on the left and carry the&nbsp;1 to it.)


Perhaps it was overly complicated to say that we put a zero on the left-end of the number and then increase it; I could have just said that we put a one on the left end and we're done. I used the slightly longer formulation because you can keep using that when you start adding numbers bigger than one.