Solution:
$(1)$ First, prove that $0 0$. Since $a_{n+1}$ and $a_n$ are of the same sign and $a_1= \frac{1}{2}$,
$\therefore 0 < a_n$,
Furthermore: $\frac{a_{n+1}}{a_n}= \frac{1-a_n}{2-a_{n+1}} < \frac{1}{1}=1$,
$\therefore a_{n+1} < a_n$ $(n\in\mathbb{N}^*)$
$(2)$ Since $a_{n+1}^2-2a_{n+1}=a_n^2-2a_n+a_n$ $(n\in\mathbb{N}^*)$,
the sum of the first $n$ terms of the sequence $\{a_n\}$ is $S_n=(a_{n+1}^2-2a_{n+1})-(a_1^2-2a_1)=a_{n+1}^2-2a_{n+1}+ \frac{3}{4}$.
From $(1)$, we have $2a_{n+1}-a_n=a_{n+1}^2-a_n^2 < 0$.
$\therefore a_n\leqslant \frac{1}{2^n}$, $\frac{3}{4}- \frac{1}{2^n} < S_n= a_{n+1}^2-2a_{n+1}+ \frac{3}{4} < \frac{3}{4}$.
$(3)$ Given $a_{n+1}^2-2a_{n+1}=a_n^2-a_n$ $(n\in\mathbb{N}^*)$,
which means $(2-a_{n+1})a_{n+1}=(1-a_n)a_n$, and $a_1= \frac{1}{2}$,
Since $b_n= \frac{1}{a_{n+1}}- \frac{2}{a_n}$,
we get: $b_1=2 \sqrt{3}$.
$\therefore b_n= \frac{1}{a_{n+1}}- \frac{2}{a_n}= \frac{1}{1-a_n}- \frac{1}{2-a_{n+1}}$.
Next, we prove the monotonicity of $\{b_n\}$:
Then $b_{n+1}-b_n= \frac{2}{1-a_{n+1}}- \frac{1}{2-a_{n+2}}- \frac{2}{1-a_n}+ \frac{1}{1-a_{n+1}}= \frac{-2(a_n-a_{n+1})}{(1-a_{n+1})(1-a_n)}+ \frac{a_{n+1}-a_{n+2}}{(2-a_{n+2})(2-a_{n+1})}$
From $(1)$, $a_{n+1} < a_n$ $(n\in\mathbb{N}^*)$ is monotonically decreasing.
Thus $b_{n+1}-b_n < 0$.
When $n=1$, the value of $b_n$ is at its maximum, hence $b_n\leqslant 2 \sqrt{3}$.
Therefore, the final answers are $\boxed{a_{n+1} < a_n}$, $\boxed{\frac{3}{4}- \frac{1}{2^n} < S_n < \frac{3}{4}}$, and $\boxed{b_n\leqslant 2 \sqrt{3}}$.