A periodic in x, Lipschitz in y, ODE with a periodic initial condition must have a periodic solution

Question

I'm trying to check if my reasoning is correct as this question was a bit different than what I previously encountered :

Question

Suppose :

1. The well-posed ODE is periodic in $x$ with period $T$ : $\forall x \in \Re : y'(x) = F(x,y) = F(x+T,y)$
2. The initial condition is also periodic with period $T$ : $y(a) = y(a+T) = Y$

If $u(x)$ is the unique solution, then $\forall x \in \Re :u(x) = u(x+T)$

Working

Since the ODE is well-posed then there exists a unique solution, call it $u(x)$.

Another way to write the given ODE is : $$u(x) = Y + \int_a^x F(x,u) dx$$

Shift $x$ by $T$ : $z = x + T$ and hence $F(z,u) = F(x,u)$ and $u(a)$ translates to (and is equal to) $u(a-T)$.

Therefore $dz = dx$ which by substitution gives : $$u(x) = u(z-T) = Y + \int_{a+T}^{x+T} F(z,u)dz = Y + \int_{a+T}^{x+T} F(x,u)dx$$ $$\therefore u(x) = Y + \int_{a+T}^{x+T} F(x,u)dx = Y + \int_a^x F(x,u) dx$$

Clearly $u(x)$ is periodic with period $T$.

Comment

I'm not entirely sure this is correct throughout, mainly how I handled the initial point. Is this idea correct or salvageable? Thank you for your time.