I suppose you want uniqueness assuming some initial condition.

Even when $F$ is continuous and $X$ is $C^1$ you don't have uniqueness. For $t\geq 0$ $X(t)=0$ and $X(t)=t^2$ are both solution of $X'=2\sqrt{X}$.

Maybe with stronger assumptions on $F$ you can do something assuming only that $X$ is Lipschitz but I'm not sure. Without the Lipschitz condition you don't have uniqueness assuming that the differential equation is satisfied a.e., Cantor stair is a counterexample.

I suppose you want uniqueness assuming some initial condition.

Even when $F$ is continuous and $X$ is $C^1$ you don't have uniqueness. For $t\geq 0$ $X(t)=0$ and $X(t)=t^2$ are both solution of $X'=2\sqrt{X}$.

Maybe with stronger assumptions on $F$ you can do something assuming only that $X$ is Lipschitz but I'm not sure. Without the Lipschitz condition you don't have uniqueness assuming that the differential equation is satisfied a.e., Cantor stair is a counterexample.

Edit : The uniqueness statement you want is true when $F$ is Lipschitz. In this case, with your assumptions, $X$ has a right derivative which is continuous. From that it's not too complicated to show that $X$ has a derivative everywhere which is continuous. And then the classical Cauchy-Lipschitz applies.

I suppose you want uniqueness assuming some initial condition.

Even when $F$ is continuous and $X$ is $C^1$ you don't have uniqueness. For $t\geq 0$ $X(t)=0$ and $X(t)=t^2$ are both solution of $X'=2\sqrt{X}$.

Maybe with stronger assumptions on $F$ you can do something but I'm not sure.

I suppose you want uniqueness assuming some initial condition.

Even when $F$ is continuous and $X$ is $C^1$ you don't have uniqueness. For $t\geq 0$ $X(t)=0$ and $X(t)=t^2$ are both solution of $X'=2\sqrt{X}$.

Maybe with stronger assumptions on $F$ you can do something assuming only that $X$ is Lipschitz but I'm not sure. Without the Lipschitz condition you don't have uniqueness assuming that the differential equation is satisfied a.e., Cantor stair is a counterexample.