If $j$ is a non-negative integer then $10^j$ is $1$ more than a multiple of $9$ so let $10^j-1=9V(j)$ where $0\leq V(j)\in \Bbb Z.$ E.g. $10^6=1+999999=1+(9)(111111)=1+9V(6).$
For $10\leq x\in \Bbb Z^+$ let $S(x)$ be the sum of the digits of $x$ and let $f(x)=x-S(x).$ Then $f(x)<x.$
And $f(x)$ is divisible by $9.$ Because if the digit-sequence for $x$ is $(x_n,...,x_0)$ then $$f(x)=x-S(x)=(x_n\cdot 10^n+...+x_0\cdot 10^0)-(x_n+...+x_0)=$$ $$=x_n(10^n-1)+...+x_0(10^0-1)=$$ $$=x_n\cdot 9V(n)+...+x_0\cdot 9V(0)=$$ $$=(9)(x_n V(n)+...x_0 V(0))\quad \bullet$$ Now if $x\geq 10$ then (i) $n \geq 1$ and $x_n\geq 1$ and $V(n)\geq V(1)=1,$ and (ii) no $x_j V(j)$ is negative whenever $0\leq j< n,$ so by $\bullet $ we have $$f(x)\geq (9)(x_nV(n))\geq 9\quad \star$$
So the sequence $x,f(x), f((f(x)),...$ (in a manner of speaking, as this sequence might have only the 2 terms $x$ and $ f(x)$)..... is a strictly decreasing sequence in which every term, except possibly the first term, is a multiple of $9$. This sequence must reach a value less than $10$ where it must stop. But the last value is $f(y)$ where $y$ is the second-last value, with $y\geq 10$. So by $\star$ we have $f(y)\geq 9.$ That is, $10>f(y)\geq 9$.
Remark. The sequence must stop when it reaches a value less than $10$ because we deliberately keep values less than $10$ out of the domain of the function $f$.