If $f(y, t)$ is continuous in a rectangle $D = \{0 \le t - t_0 \le a$, $|y - y_0| \le b \}$ then the initial value problem $y' = f(y, t)$ with $y(t_0) = y_0$ has at least a solution in $[t_0, t_0 + \alpha]$, $\alpha = \min(a, b/M)$, $M=\max_D|f|$, by the Peano existence theorem.
On the book by Hartman there is an exercise that asks to prove this theorem by building a sequence of partitions $\Sigma_n = \{t_0,t_1,t_2,\dots,t_{n-1}, t_n=t_0 +\alpha\}$ of $[t_0, t_0 + \alpha]$, whose granularity $\max_k|t_{k+1}-t_k|=\delta_n \rightarrow 0$ as $n\rightarrow \infty$. On each partition a function is defined to be piecewise linear and such that: $$ y_n(t_0) = y_0 \\ y_n(t) = y(t_k) + f(y(t_k), t_k) (t - t_k), \quad t \in [t_k, t_{k+1}], \ k=0,\dots, n-1 $$
All functions are $C^0[t_0, t_0+\alpha]$, uniformly bounded in the $C^0$-norm ($|y_n(t) - y_0| \le b$) and also equi-continuous as $|y_n(t) - y_n(s)| \le M |t - s|$ by definition. By the Ascoli-Arzelà theorem we have (up to a subsequence) a uniform limit $y \in C^0$.
I think the equicontinuity comes from the fact that: $$ y_n(t) = y_0 + \int_{t_0}^{t} \sum_{k=0}^{n-1} f(y_n(t_k), t_k)1_{(t_k, t_{k+1})}(s)ds $$
I am left to show that $y$ satisfies the following integral equation: $$ y(t) = y_0 + \int_{t_0}^{t} f(y(s), s) ds, \quad t \in [t_0, t_0 +\alpha] $$
Defining: $$ g_n(t) = \sum_{k=0}^{n-1} f(y_n(t_k), t_k)1_{(t_k, t_{k+1})}(t) $$
I tried to prove that $g_n(t) \rightarrow f(y(t), t)$ uniformly in order to be able to exchange the limit and the integral. Given a fixed $\varepsilon > 0$: $$ |g_n(t) - f(y(t),t)| \le |g_n(t) - f(y_n(t), t)| + |f(y(t), t) - f(y_n(t), t)| $$
The second term is $\le \frac{\varepsilon}{2}$, for all $t$, because of the uniform convergence of $y_n$ and the uniform continuity of $f$. But how to treat the first term?