This is not an answer but a (long) comment to Did's answer. In a nutshell: I think that we need to add a Lipschitz condition on $F$ (assumption 2 below). Did probably assumed it implicitly.
Here's a possible way to argue. Assume that:
- for all $s\in[0, T]$, one has that $x\le y\Rightarrow F(s, x)\le F(s, y)$;
- there exists a constant $L>0$ such that $\lvert F(s, x)-F(s, y)|\le L\lvert x-y\rvert$.$^{[1]}$
Let
$u(t)\le c+\int_0^tF(s, u(s))\, ds,\qquad w(t)=c+\int_0^tF(s, w(s))\, ds.$
We claim that $\tag{1}u(t)\le w(t).$ Proof. Consider the set $ X=\left\{t\in [0, T]\ :\ u(t)-w(t)>0\right\}.$ Assume by contradiction that $X$ is nonempty and set $ t_0=\inf X.$ Since $u$ is a subsolution we have that $t_0>0$. By assumption 1 we have that $ F(s, u(s))-F(s, w(s))\le 0\qquad \forall s\in[0, t_0]. $ So for $t\in X$ we have by assumption 2 \begin{equation} \begin{split} 0$0. $\square$
I don't know if condition 2 can be weakened, but surely it cannot be dropped altogether. One must have at least a uniqueness result for the integral equation, and condition 2 gives such a result (that's the standard Picard's existence and uniqueness theorem).
For example, consider this problem:
$w(t)=\int_0^t \big(w(s)\big)^{1/3}\, ds.$
We know that there exist more than one solution to this equation, and of course every solution is a subsolution. If our claim were true in this case we would have for a pair of distinct solutions $w_1, w_2$ the inequalities $w_1(t) \le w_2(t)$ and $w_2(t)\le w_1(t)$, that is, $w_1(t)=w_2(t)$, a contradiction.
An open question remains, and it it necessity of the monotonicity condition on $F(s, \cdot)$. This is intuitively reasonable, but should be proved.
$^{[1]}$ It is enough to assume those conditions for almost all values of $s$.