Proving $((a\bmod n) (b\bmod n))\bmod n = ab\bmod n $

Note that by the Chinese Remainder Theorem, for any $k\gt 0$, and for any integers $a$ and $b$, there exists an integer $x$ such that $x\bmod k = a\bmod k$ and $x\bmod (k+1) = b\bmod (k+1)$. That is: the remainders of $x$ modulo $k$ and modulo $k+1$ are completely unrelated. So I do not see how you are going to be able to leverage "knowing" the result modulo $k$ into a proof of the result modulo $k+1$, unless you simply prove it directly modulo $k+1$.

So it is really simpler to show that the result holds modulo $n$ directly, for any $n\gt 0$.

Remember that $x\bmod n = r$ if and only if $0\leq r\lt n$ and $x-r$ is a multiple of $n$.

So first show that $ab - (a\bmod n)(b\bmod n)$ is a multiple of $n$. For example, if $a\bmod n = r$ and $b\bmod n = s$, then $a-r$ and $b-s$ are both multiples of $n$; then $(a-r)b$ is a multiple of $n$, and $r(b-s)$ is a multiple of $n$, so...

Let $c = a\bmod n$, true iff $a = jn +c$ for some integer $j$. Similarly $d = b\bmod n$ iff $b = kn +d$ for some integer $k$.

So,

$ab\bmod n = [(jn+c) \times (kn+d)] \bmod n $

$= [(jkn+jd+kc)n + c \times d] \bmod n $

$ = c \times d \bmod n $

$= [(a\bmod n)\times (b\bmod n)]\bmod n$,

which is what you wanted to prove.