Let $f:B_r(x_0)\to\mathbb{R}^n$ be continuous. Prove there always exists a local solution $x:[0,\delta)\to\mathbb{R}^n$ satisfying $$x(0)=x_0, \hspace{1cm} x'(t)=f(x(t)), \quad \forall t \in (0,\delta),$$ by approximating $f$ uniformly by $C^{\infty}$ functions.
I am looking for proof verification of the following and any suggestions for improvement. I do not feel confident with this proof because I did not use Ascoli-Arzela, which is used in the typical proof of Peano's Existence Theorem.
Consider, instead of the open ball $B_r(x_0)$, a closed ball contained in it (since we're showing a local solution, this makes no difference), say $\overline{B_r(x_0)}$. Consider polynomials $P_n(x)$ which converge uniformly to $f$ on the closed ball. These exist because of the Weierstrass' theorem. Now the problems $$\begin{cases} x_n'(t)=P_n(x_n(t))\\ x_n(0)=x_0\end{cases}$$ have a unique local solution in $B_r(x_0)$, since $P_n$ is a polynomial, hence locally Lipschitz on the open ball. Call this solution $x_n$.
Moreover, we may extract a convergent subsequence of $x_n$ such that the function $P_n$ has a uniform limit $f$, and therefore the functions $t \mapsto x_n'(t)$ have a uniform limit, which we shall denote by $g(t)$.
Now we have a sequence of functions $x_n$ which take the same value at $0$, which are $C^1(B_r(x_0))$ and whose derivatives converge uniformly to a function $g$. Therefore $\exists \lim x_n$ and it is differentiable in a neighborhood of $0$ and its derivative is $g$ because:
If $\exists a.f_n(a)\to L$ and $f'_n\to g$ uniformly then $f_n$ has a limit $f$ satisfying $f(a)=L$, $f'=g$.
In other words, there is some local solution $x:[0,\delta)\to\mathbb{R}^n$ satisfying $x(0)=x_0$ where $x'(t) = f(x(t))$ for all $t\in[0,\delta)$.