Does the implicit function theorem imply Peano existence theorem In The implicit function theorem written by Krantz & Parks, it's said that the implicit function theorem implies the following existence theorem of ODE:

Theorem 4.1.1 If $F(t,x)$, $(t,x)\in\mathbb R\times\mathbb R^N$, is continuous in the $(N+1)$-dimensional region $(t_0-a,t_0+a)\times B(x_0,r)$, then there exists a solution $x(t)$ of
  $$\frac{dx}{dt}=F(t,x),\qquad x(t_0)=x_0$$
  defined over an interval $(t_0-h,t_0+h)$.

It's Peano existence theorem, I think. However, there seems a gap in their proof. WLOG, suppose $t_0=0$. They constructed $\mathcal H\colon[0,1]\times\mathcal B_1\to\mathcal B_0\times\mathbb R$, where $\mathcal B_0$ is the space of bounded continuous $\mathbb R^N$-valued functions on $(-a,a)$ normed canonically, and $\mathcal B_1$ is the space of bounded continuously differentiable $\mathbb R^N$-valued functions on $(-a,a)$ that also have a bounded derivative, normed canonically by $\sup\lvert f\rvert+\sup\lvert\dot f\rvert$, as follows:
$$\mathcal H[\alpha,X(\tau)]=[X'(\tau)-\alpha F(\alpha\tau,X(\tau)),X(0)-x_0]$$.
Note that $\mathcal H[0,x_0]=[0,0]$, where $x_0$ on the left side denotes the constant function. Then they claim that the existence theorem follows from the implicit function theorem. However, under the only condition that $F$ is continuous, there's no evidence that $\mathcal H$ is partially differentiable with respect to $X$ for $\alpha\in(0,x_0)$.
Can we fix the preceding proof in some extent?
PS: I posted the question not only because I want to comprehend such a proof, but also want to understand the relation between ODE and implicit functions. It seems certain that such a proof cannot be alright, since the canonical implicit function theorem is also a uniqueness theorem, which implies the local uniqueness of a solution of ODE. However, I want to know how to fix it. I doubt it might rely on a more general implicit function theorem.
 A: Observe that we have Fréchet-differentiability in the second argument at the point where you want your implicit function. Indeed, the function $T: Y \mapsto (Y', Y(0))$ is linear, and we have
$$
\lim_{Y \to 0} \frac{\mathcal H(0, X + Y) - (\mathcal H(0, X) + T(Y))}{\|Y\|_1} = \lim_{Y \to 0} \frac{(0, 0)}{\|Y\|_1} = 0,
$$
which is why $d_2 \mathcal H(0, x_0)$ exists and is equal to $T$. Note that $T$ is invertible, since $Y(0)$ determines the integration constant. Furthermore, we can obtain the following modified version of theorem 3.4.10:
Let $X, Y, Z$ be Banach spaces. Let $U \times V \subseteq X \times Y$, $g : U \times V \to Z$ and let $(x, y) \in U \times V$ such that $d_2(x, \cdot)$ exists and $G$ is continuous and $G(x, y) = 0$, and further such that there exists $W \subseteq V$ such that $y \in W$ and $G(z, w) \to G(x, w)$ as $z \to x$ uniformly for $w \in W$. Then there exist $M \subseteq U$ and $N \subseteq W$ such that for each $\xi \in X$, there exists a unique $\eta \in M$ such that $G(\xi, \eta) = 0$ and the thereby defined function is continuous.
In the proof, we replace the mean value theorem in part by the uniform convergence from the assumption, where we choose $N$ small enough such that $\|d_2 L(x, \cdot)\| < 1/4$.
From this theorem, we obtain Peano's existence theorem as in the given proof.
Note also that uniqueness does not follow, since the IFT only gives uniqueness locally around the constant function.
