Suppose we have a Cauchy problem
$$x'=f(x(t),t),\quad x(t_0)=C,$$
where $f$ is Lipschitz continuous in its first argument and continuous in its second argument. By Picard theorem, there must exist a unique solution within some interval $t\in[t_0-\varepsilon,t_0+\varepsilon]$. Now, on this interval, construct a sequence of functions $(x_n)$ by recursive formula
$$x_0(t):=C,\quad x_{n+1}(t):=\int_{t_0}^tf(x_n(s),s)\,\mathrm ds+C.$$
It is clear that $x_{n+1}'(t)=f(x_n(t),t)$, and if we take the limit $n\to\infty$, we can see $\lim_{n\to\infty}x_n$ is the solution to the original ODE if the sequence $(x_n')$ converges uniformly on the interval we consider.
Because of the uniform convergence condition, we can potentially get a wrong solution by this procedure. Is it guaranteed that $\lim x_n$ exists? Can we construct $f$ and $C$ such that $(x_n')$ fails to converge uniformly, and $\lim x_n$ exists but is a wrong solution? If we cannot, then is it guaranteed that $\lim x_n$ is a solution as long as it exists? Can we find the sufficient and necessary condition for $\lim x_n$ to exist and be the solution?
The hypothesis of Picard–Lindelöf theorem are sufficient to ensure that the sequence $\{x_n\}$ converges uniformly. See here for a proof of Picard–Lindelöf theorem that proves this uniform convergence as a by-product.