Theorem: Given a continuous function $f:(a,b)\times (c,d)\to \mathbb R$, the problem $$\begin{cases} y'(t)=f(t,y(t))\\ y(t_0)=y_0\end{cases}$$ has a local solution in some neighbourhood of a given point $t_0$.
A "standard" proof of Peano existence theorem makes use of the Ascoli-Arzelà theorem. I suspect that the following proof, which doesn't, is therefore wrong. But where?
Consider, instead of the open rectangle $(a,b)\times (c,d)$, a closed rectangle contained in it (since the thesis is local, this makes no difference), say $[a',b']\times[c',d']$. Consider polynomials $P_n(t,y)$ which converge uniformly to $f$ on the closed rectangle. These exist because of the Weiertrass theorem. Now the problems $$\begin{cases} y_n'(t)=P_n(t,y_n(t))\\ y_n(t_0)=y_0\end{cases}$$ have a unique global solution in $[a,b]$, since $P_n$ is a polynomial, hence globally Lipschitz on the rectangle (recall that globally Lipschitz implies global existence by gluing local solutions).
Now we say that the function $P_n$ has an uniform limit $f$, and therefore the functions $t\to y_n'(t)$ have an uniform limit, which we shall denote by $g(t)$. (This is the point which I am less sure of.)
Now we have a sequence of functions $y_n$ which take the same value in $t_0$, which are $C^1([a',b;])$ and whose derivatives converge uniformly to a function $g$. Therefore a limit for $y_n$ exists, is differentiable in a neighbourhood of $t_0$ and its derivative is $g$. (This is a theorem of elementary analysis: $f_n(a)\to l$ for a given $a$, $f'_n\to g$ uniformly, then $f_n$ has a limit $f$ which satisfies $f(a)=l$, $f'=g$.)
Thank you in advance for helping me finding the (possible) error.