Just another way.
Clearly $9\vert (10-1)$. Suppose that $9\vert (10^n - 1)$, $n\gt 1$. Then
$$10^{n+1}-1 =10\cdot 10^{n}-1=9\cdot 10^n + (10^n - 1)$$
thus $9\vert (10^{n+1}-1)$. Therefore $9\vert (10^n-1)$ for any $n\in\mathbb{N}$.
Now take an integer
$$a=\sum_{i=0}^n a_i10^{i},$$
where $a_i\in \{0,1,\ldots,9\}$. Then
$$a-\sum_{i=0}^na_i=\sum_{i=0}^n a_i10^{i}-\sum_{i=0}^na_i=\sum_{i=1}^n a_i(10^{i}-1),$$
therefore, by the established above
$$a\equiv \sum_{i=0}^na_i \pmod 9$$
We write $n=\overline{abcd}$, and as in your question $a_4$ to $a_1$ are the numbers from large to small.
$$a = a_4-a_1$$
$$b = a_3-a_2-1$$
$$c = a_2-a_3+9$$
$$d = a_1-a_4+10$$
We know that $a_4 \geq a_3 > a_2 \geq a_1$. We also know $a+d=10$,$b+c=8$.
We also know $a > b$.
We also know $c \geq d-1$, with equality iff $a_4 = a_3 > a_2 = a_1$, in that case, we can write $n=\overline{eeff}-\overline{ffee}=\overline{(e-f)(e-f)(f-e)(f-e)}$ if $n$ is Kaprekar, but this doesn't statistify $a>b$ and thus it can't be Kaprekar. Therefore $c \geq d$.
We therefore have 6 possibilities for $n=\overline{abcd}$:
It can be $\overline{a_4a_2a_3a_1}$, $\overline{a_3a_2a_4a_1}$, $\overline{a_4a_1a_3a_2}$, $\overline{a_3a_1a_4a_2}$, $\overline{a_2a_1a_4a_3}$ or $\overline{a_4a_3a_2a_1}$.
Case 1. $\overline{a_4a_2a_3a_1}$. In this case we have $a_4=a_4-a_1$, so $a_1=0$. But $a+d=a_1+a_4=10$, so $a_4=10$. Contradiction.
Case 2. $\overline{a_3a_2a_4a_1}$. In this case we have $a_1=a_1-a_4+10$, so $a_4=10$. Contradiction.
Case 3 and 6. $\overline{a_4a_1a_3a_2}$. See case 1.
Case 4. $\overline{a_3a_1a_4a_2}$. This is the case 6174 is in. We have the system:
$$a_3 = a_4-a_1$$
$$a_1 = a_3-a_2-1$$
$$a_4 = a_2-a_3+9$$
$$a_2 = a_1-a_4+10$$
Substitute first in the latter three:
$$a_1 = a_4-a_1-a_2-1$$
$$a_4 = a_2-a_4+a_1+9$$
$$a_2 = a_1-a_4+10$$
Substitute last in the former two:
$$a_1 = a_4-a_1-a_1+a_4-10-1$$
$$a_4 = a_1-a_4+10-a_4+a_1+9$$
Rewrite:
$$3a_1 = 2a_4-11$$
$$3a_4 = 2a_1+19$$
Two times first and three times second added gives:
$$6a_1+9a_4 = 4a_4-22+6a_1+57$$
$$5a_4 =35$$
$$a_4 =7$$
This gives $3a_1 = 14-11=3$ thus $a_1=1$. Filling this in in the first and last original equation gives $a_3=6$ and $a_2=4$. This gives $n=6174$ as Kaprekar number.
Case 5:
$$a_2 = a_4-a_1$$
$$a_1 = a_3-a_2-1$$
$$a_4 = a_2-a_3+9$$
$$a_3 = a_1-a_4+10$$
Substitute first in the second:
$$a_1 = a_3+a_1-a_4-1$$
$$0 = a_3-a_4-1$$
$$a_4+1 = a_3$$
This contradicts the fact that $a_4 \geq a_3$.
This completes the proof. Phew.
Best Answer
Writing $9n=10n-n$ is a good start. The idea then is to note that, since the digits are strictly increasing, it is easy to see how that subtraction must turn out. Indeed, in most places you are subtracting a number from one which is strictly greater. More precisely, other than in the units and the tens place we are subtracting $a_i$ from $a_{i+1}$, thereby producing the gap between $a_{i+1}$ and $a_i$. (using your notation) In the units place we are, of course, subtracting the final digit from $0$ which will require a simple carry.
Consider the gap sequence, pretending that the number begins with a $0$. Thus, if your starting number were $24578$ the gap sequence would be $\{2,2,1,2,1\}$. We remark that the partial sums of the gaps return the original digits. In particular, the sum of all the gaps is the final digit...let's call it $d$.
Then the digits in $9n=10n-n$ are precisely the digits in the gap sequence, save that the final one is lowered by $1$ and we append $10-d$. All that remains is to remark that the sum of the digits in that string is precisely $d-1+10-d=9$ and we are done.
Example: sticking with $n=24578$ we get $9n=221202$ as claimed.
Remark: the lowering of the penultimate gap and the appending of $10-d$ are the result of the obvious carry, when you try to subtract $d$ from $0$. Of course, if $n=0$ at the start, then there is no carry. In all other cases, there is always a carry at that stage (and no other).