I'm currently taking an elementary number theory course, and not quite sure if this proof suffices for the given problem statement.
Here is my proposed proof:
Let $d$ be an $n$-digit natural number.
Let there be two of the digits $a,b$ for $d$ where $a = a*10^x$, $b = b*10^y$, $0\leq x\neq y \leq n - 1$.
Assume $x > y$ to always achieve a non-negative difference.
When swapping the digits a and b, we have the following difference between the two quantities:
$(a * 10^x + b*10^y) - (b*10^x + a*10^y) \leftrightarrow $ $a(10^x - 10^y) + b(10^y - 10^x) \leftrightarrow (a - b)(10^x - 10^y)$
Since $10 \equiv 1(\mod 9)$, the above statement evaluates to 0, thus showing $9 | 0$, which means the difference between the correct amount and the amount with the two digits switched is always divisible by 9.
I would drop the $x > y$ requirement. It's clutter, in my opinion. It suffices that $x \neq y$.
Take for example $1729$. Switch two digits to get $1927$. Then $1927 - 1729 = 198$. But $1729 - 1927 = -198$. Taking your assertion
$(a * 10^x + b*10^y) - (b*10^x + a*10^y) \leftrightarrow $ $a(10^x - 10^y) + b(10^y - 10^x) \leftrightarrow (a - b)(10^x - 10^y)$
and plugging in $7$ and $9$ both ways, we get
$$(700 + 9) - (900 + 7) \leftrightarrow (7 - 9)(100 - 1)$$
and
$$(900 + 7) - (700 + 9) \leftrightarrow (9 - 7)(1 - 100)$$
Then, ignoring signs, we arrive at the same result.