This is a comment to complement @S.Maths response.

Suppose we show that $$\|u(t)-u(t')\|_X\leq C|t-t'| \hbox{ for all } t,t' \in [0,T]$$ and some $C>0$.

Since $F$ is locally Lipschitz, there is $L_M>0$ such that $$\|F(v)-F(w)\|\leq L_M\|v-w\|$$ for all $v,w \in X$ such that $\|v\| \leq M$ and $\|w\| \leq M$.

Taking into account the continuity of the norm, the function $\|u(\cdot)\|:[0,T] \rightarrow \mathbb{R}$ is continuous on the compact set $[0,T]$.

Therefore, there is $M>0$ such that $\|u(t)\| \leq M$ for all $t \in [0,T]$. Thus, $$\|F(u(t))-F(u(t'))\|\leq L_M\|u(t)-u(t')\|\leq CL_M|t-t'|$$ for all $t,t' \in [0,T]$, which proves that $t \in [0,T] \mapsto F(u(t))$ is Lipschitz continuous, no matter if $F$ is locally Lipschitz or locally Lipschitz.

This is a comment to complement @S.Maths response.

Suppose we show that $$\|u(t)-u(t')\|_X\leq C|t-t'| \hbox{ for all } t,t' \in [0,T]$$ and some $C>0$.

Since $F$ is locally Lipschitz, there is $L_M>0$ such that $$\|F(v)-F(w)\|\leq L_M\|v-w\|$$ for all $v,w \in X$ such that $\|v\| \leq M$ and $\|w\| \leq M$.

Taking into account the continuity of the norm, the function $\|u(\cdot)\|:[0,T] \rightarrow \mathbb{R}$ is continuous on the compact set $[0,T]$.

Therefore, there is $M>0$ such that $\|u(t)\| \leq M$ for all $t \in [0,T]$. Thus, $$\|F(u(t))-F(u(t'))\|\leq L_M\|u(t)-u(t')\|\leq CL_M|t-t'|$$ for all $t,t' \in [0,T]$, which proves that $t \in [0,T] \mapsto F(u(t))$ is Lipschitz continuous, no matter if $F$ is locally Lipschitz or globally Lipschitz.