I have an ODE $$u' = f(t,u) = \int_0^t k(u(s)) \text{ d}s$$ with an initial data $u_0$. Here, $k \in C_b(\mathbb{R})$ is a continuous and bounded function such that $|k(x)| \leq C$ for all $x\in \mathbb{R}$.
I want to apply the Cauchy-Peano theorem to this ODE to show that there exists a $C^1$ solution $u$.
Let $u$ be fix. Then $t \mapsto f(t,u)$ is continuous, see this question.
Let $t$ be fix. And let $u_n(x) \to u(x)$ for every $x \in \mathbb{R}$ (almost every $x$ would be enough). Then, since $k$ is continuous and bounded, we also have $k(u_n(x)) \to k(u(x))$ for almost every $x$ and $|k(u_n(x))| \leq C$. Therefore, by the Lebesgue dominated convergence theorem, we have
$$f(t,u_n)=\int_0^t k(u_n(s)) \text{ d}s \longrightarrow \int_0^t k(u(s)) \text{ d}s=f(t,u),$$
and $u \mapsto f(t,u)$ is continuous for fixed $t$.
Therefore, Cauchy-Peano gives me the existence of a $C^1$ solution to the above ODE.
Is this correct?
Your argument does not give you the existence of a solution, only that if an approximating and converging sequence exists then the limit satisfies (parts of) the original equation.
But you can easily transform your equation into an ordinary ODE by differentiating once more, $$ u''=k(u), ~~ u(0)=u_0,~ u'(0)=0. $$ To that or its first order version $$ u'=v\\ v'=k(u) $$ you can now apply the usual Peano theorem and conclude the existence of local solutions.