Consider the initial value problem $$\dot x(t) = F(t,x), \quad t \in (0,T)$$ with given initial datum $$x(0) = x_0 \in \mathbb R.$$ More precisely we consider the integral equation $$x(t)=x(0)+\int_0^t F(s,x(s))ds.$$
$F$ may be discontinuous, but let us assume that $$0 < m < F(t,x) < M.$$
The common counter-examples to uniqueness (or existence) of ODEs (or their associated integral equations) seem to rely on $F$ switching sign, or being close to $0$, and my intuition is that the lower bound $m<F(x)$ should imply existence of a unique solution.
Question 1: Is it true that there exist a solution under the assumptions above?
Question 2: Can we also prove uniqueness?
The answer to both questions is no. The most general existence theorem I know is Carathéodory's existence theorem. It requires that $F(t,x)$ is continuous in $x$ for each fixed $t$, measurable in $t$ for each fixed $x$ and a boundedness condition similar to the one of the question (but $m$ may be negative). Without a minimum of conditions of $F$, the function $s\to F(s,x(s))$ may not even be measurable.
The positivity of $m$ does not imply uniqueness as a simple variant of the classical counterexample shows. Consider $$x'=1+2\sqrt{|x-t|}, x(0)=0.$$ The transformation $z=x-t$ shows that it is quivalent to $z'=2\sqrt z, z(0)=0$. Two solutions are, for example, $x_1(t)=t+t^2$ for $t\geq0$, $x_1(t)=t$ otherwise and $x_2(t)=t$ for all $t$. Actually, the initial value problem, as the classical one, has a two-parameter family of solutions. So clearly, $F(t,x)\geq1$ for all $t,x$, but there is no uniqueness.