The problem here is to prove that $f$ is strictly decreasing on $[a, b]$ if it is differentiable on $[a, b]$, $f'$ is not identically zero for any non-degenerate subinterval of $[a, b]$, and $f'(x)\le0$ for all $x\in[a, b]$. I have already proven the converse for if $f$ is strictly increasing when $f'(x)\ge0$, but on this proof I think the non-degenerate interval is tripping me up. Here is what I have so far (using MVT):
We will assume that $f$ is differentiable, with $f'(x)\le 0$ for all $x$ in the interval. We also will assume that $f$ is not strictly decreasing. Therefore, there exists $x_1<x_2\in [a, b]$ such that $f(x_1)\le f(x_2)$, and combining these inequalities gives us that $\frac{f(x_2)-f(x_1)}{x_2-x_1}\ge 0$.
Because $f$ is differentiable on $[a,b]$, it is also continuous on $[x_1, x_2]$ and differentiable on $(x_1, x_2)$. Therefore, by the MVT, there exists at least one value of $c\in(x_1, x_2)$ such that:
$f'(c)=\frac{f(x_2)-f(x_1)}{x_2-x_1}$.
By this assumption, $f'(c)\le 0$
$\implies\frac{f(x_2)-f(x_1)}{x_2-x_1}=0$
$\implies f(x_2)-f(x_1)=0$
$\implies f(x_2)=f(x_1)$.
In the proof of strictly increasing, I then went on to do a proof by contradiction, assuming $f$ is not constant on $[a, b]$ and showing that that results in a contradiction. What should I do in this case?
First, show that $f$ is (weakly) decreasing. Take any points $x_1<x_2$ in the interval $[a,b]$. By the mean value theorem, there exists a $c\in(x_1,x_2)$ such that $$\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(c)\leq 0 $$ Hence, $f(x_2)\leq f(x_1)$, and $f$ is decreasing in $[a,b]$. If you assume that $f$ is not strictly decreasing, there exist $x_3<x_4$ in $[a,b]$ such that $f(x_3)=f(x_4)$. But $f$ is decreasing in $[x_3,x_4]$. Thus, it's constant in that interval, and its derivative there is $0$, a contradiction.