May I inquire as to why the result is always 9?

It's actually quite intuitive!

First, let us state that to have a two digit integer whose digits add to nine is equivalent to saying that the two digit integer is divisible by nine:

A two digit integer can be written ab (eg in 23, a=2 and b=3)

Then ab=10a+b

ab = a+b (mod 9)

So the number ab is divisible by 9 if the sum of the digits is

divisible by nine.

Now, consider an arbitrary two-digit integer, nm.

Then nm=10n+m, as above

And nm-(n+m)=10n+m-n-m=9n

Thus, the result of subtracting the sum of the digits from the original integer is a multiple of nine; equivalently, the sum of the digits of the result will be a multiple of nine.

How do we know the sum is 9, and not a higher multiple of 9? Because the only integer whose digits sum to a multiple of nine which is greater than nine is 99, an impossible two-digit integer to obtain after performing our subtraction.

Cool trick!