Your problem comes when you assume that there were $3$ taken from your last set. It's quite possible that the second or third choice was a $0$, so that you in fact have only eliminated $2$ options from the last set.
Instead, choose the last digit first ($5$ choices), then the first digit ($8$ choices, since we can't have picked $0$ for the last digit), then deal with the second and third digit ($8$ and $7$ choices, in some order). That will get you $5\cdot 8\cdot 8\cdot 7=2240$, as desired.
Hint: There are two types of $4$ digit odd integers with all digits different: (i) First digit is even ($2,4,6,8$) or (ii) first digit is odd. Count the Type (i) numbers, the Type (ii) numbers, and add.
Type (i): There are $4$ choices for the first digit. For each of these choices there are $5$ choices for the last digit. For each of the choices we have made so far, there are $8$ choices for the second digit, and for each such choice $7$ choices for the third, a total of $(4)(5)(8)(7)$.
Now it's your turn: make a similar calculation for Type (ii).
The analysis for the even numbers will be roughly similar. The two cases are (i) last digit is $0$ and (ii) last digit is not $0$.
Remark: The tricky thing is that the first digit cannot be $0$. A slightly nicer approach, I think, is to first allow $0$ as a first digit, and then take away the "numbers" with first digit $0$, which we should not have counted.
So for odd numbers, if we allow $0$ as a first digit, we can choose the last digit in $5$ ways, and then for the rest we have $(9)(8)(7)$ choices, for a total of $(5)(9)(8)(7)$.
Now if we have $0$ as the first digit, the rest can be filled in $(5)(8)(7)$ ways.
So our answer is $(5)(9)(8)(7)-(5)(8)(7)=2240$.
For the evens, the same analysis gives $(5)(9)(8)(7)-(4)(8)(7)$.
Best Answer
You have the think the other way around.
First, let's compute the number of odd integers between 0 and 99. We can leave out 100 since it is even. Any integer between 0 and 99 has a unit digit and a tens digit, i.e. write $1,2,\ldots 9$ as $00, 01, 02,\ldots 09$. Let $U$ be the event of choosing a digit for the unit. This can be done in 5 ways (e.g. 1,3,5,7 or 9). Then let $D$ be the event of choosing a digit for the tens; this can be done in 10 ways. Since the number of $U$ that can occur doesn't depend on the number of $D$ that can occur, and vice versa, the required number is $10\cdot 5=50$.
Now back to your problem. Let $D$ and $U$ be as above. Then $U$ can occur again in 5 ways and after this, $D$ can occur in $9$ ways. This is because if $U$ happens to be 1, then $D$ should be anything than 1. Since the number of $U$ that can occur doesn't depend on the number of $D$, the multiplication rule applies and the result is $9\cdot 5=45$.