= Solution
Set $y_k(0)=x_0$ and let $M=\sup_{z\in\mathbb R^n}\lVert F(z)\rVert<\infty$. On the first interval, $y_k$ is the known constant $x_0$, so the <integral equation> defines $x_k$. Its endpoint value then defines the constant $y_k$ on the next interval, and induction over the $k$ subintervals defines the pair on all of $[0,1]$. An integral of the bounded, piecewise constant function $F\circ y_k$ is <continuous>, so every $x_k$ is continuous. In fact,
$$
\lVert x_k(t)-x_k(s)\rVert\leq M|t-s|,
$$
so the approximations are uniformly <Lipschitz continuous>.
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