Let $A_\alpha := \{ t\in \R^+_0 : \forall s\le t: |F(s) - F(0)|\le \alpha s \}$ for $\alpha > 0$. The set $A_\alpha$ is not empty because $0\in A_\alpha$.

Suppose that $t_n$ is a sequence in $A_{\alpha}$ that converges to some $t\in \R^+_0$. Since $F$ is continuous and $|F(t_n) - F(0)|\le \alpha t_n$ for all $n$, we have $|F(t) - F(0)|\le \alpha t$. Moreover, if $0\le s < t$, then there exists some $n$ such that $s < t_n\le t$, and since $t_n\in A_{\alpha}$, this implies that $|F(s)-F(0)|\le \alpha s$. Hence, $t\in A_{\alpha}$, and so $A_{\alpha}$ is closed.

On the other hand, if $t\in A_{\alpha}$, since $D_+F(t) = 0$, there exists $h_0>0$ such that $|F(t+h) - F(t)|\le \alpha h$ for all $0 < h < h_0$. Now let $t< s < t+h_0$, then we have \[ |F(s)-F(0)| \le |F(s)-F(t)| + |F(t)-F(0)|\le \alpha(s-t) + \alpha t = \alpha s. \] Moreover, for all $s\le t$ the inequality $|F(s)-F(0)|\le \alpha s$ follows from the fact that $t\in A_{\alpha}$. Thus, we conclude that $[0, t+h_0)\subseteq A_{\alpha}$, and since $t$ was arbitrary, the set $A_{\alpha}$ is open as a subset of $\R^+_0$.

Since $A_{\alpha}$ is closed, open and not empty, we conclude by connectedness of $\R_0^+$ that $A_{\alpha} = \R^+_0$ for all $\alpha > 0$, i.e. for all $\alpha > 0$ and all $s \in \R^+_0$, we have $|F(s)-F(0)|\le \alpha s$, and so $F(s) = F(0)$ for all $s \in \R^+_0$.