Let $F(t,x)$ be a function which is decreasing in $x$ for every fixed $t$.
Let $x_1,x_2$ be two solutions of the equation $\dot{x}=F(t,x)$.
Let $x_s(t)=x_1(t)-x_2(t)$.
Prove that $\forall a<b:|x_s(b)|\leq|x_s(a)|$.
I tried to write $x_i(t)=x_i(a)+\int_a^t F(s,x_i(s))ds$ for $i\in\{1,2\}$ and $t>a$. So:
$$x_s(b)=x_s(a)+\int_a^b \big[F(s,x_1(s))-F(s,x_2(s))\big]ds$$
I don't know how to use the decreasing part, it seems that applying absolute value won't help since using the triangle inequality will not give the desired bound.
Assume that $F$ is defined, continuous and nonincreasing in the second variable on $\mathbb{R} \times \mathbb{R}$.
For definiteness, let $x_1(a) > x_2(a)$. We want to prove that $$ x_1(b) - x_2(b) \le x_1(a) - x_2(b) $$ for any $b > a$ belonging to the intersection of the domains of $x_1$ and $x_2$.
Suppose to the contrary that there exist $b > a$ such that $$ x_1(b) - x_2(b) > x_1(a) - x_2(a). $$ Let $$ \theta := \inf\{ t \in [a, b]: x_1(t) - x_2(t) > x_1(a) - x_2(a)\}. $$ One has $\theta \in [a, b)$, and $x_1(\theta) - x_2(\theta) = x_1(a) - x_2(a)$. Apply the MVT to $x_1 - x_2$ on $[\theta, b]$ to obtain the existence of $\tau \in (\theta, b)$ such that $$ (x_1(b) - x_2(b)) - (x_1(\theta) - x_2(\theta)) = (x_1'(\tau) -x_2'(\tau)) \cdot (b - \theta). $$ The LHS is positive. But $x_1'(\tau) -x_2'(\tau) = F(\tau, x_1(\tau)) - F(\tau, x_2(\tau)) \le 0$, a contradiction.