Dilemma in a proof of the Implicit Function Theorem

I was studying the Implicit Function Theorem (source Diferencijalni račun funkcija više varijabli by I. Gogić, P. Pandžić and J. Tambača, pages 91-92) and I stumbled across something. First, I'm going to write the statement and proof as given in the script verbatim from Croatian:

Let $A\subseteq\Bbb R^n\times\Bbb R^m$ be an open set, $F:A\to\Bbb R^m$ of the class $C^p,p\ge 1.$ Suppose $(x^0,y^0)\in A$ satisfies $F(x^0,y^0)=0_{\Bbb R^m}$ and let $\frac{\partial F}{\partial y}(x^0,y^0)\in M_m(\Bbb R)$ be a regular matrix. Then, there is an open neighbourhood $U\subseteq\Bbb R^n$ of $x^0$ and an open neighbourhood $V\subseteq\Bbb R^m$ of $y^0$ and a unique function $f:U\to V$ of the class $C^p$ s. t. $F(x,f(x))=0_{\Bbb R^m},x\in U.$

Note $$\frac{\partial F}{\partial y}(x,y)=\begin{bmatrix}\frac{\partial F_1}{\partial y_1}(x,y)&\ldots&\frac{\partial F_1}{\partial y_m}(x,y)\\\vdots&\ddots&\vdots\\\frac{\partial F_m}{\partial y_1}(x,y)&\ldots&\frac{\partial F_m}{\partial y_m}(x,y)\end{bmatrix}$$

proof:

The idea of the proof is to apply the Inverse Function Theorem so, let's define a function $G:A\to\Bbb R^n\times\Bbb R^m$ by formula $G(x,y)=(x,F(x,y)).$ Since $F$ is of the class $C^p,G$ is also of the class $C^p$.

The Jacobian matrix of the function $G$ is given by

$\nabla G(x,y)=\begin{bmatrix}1&\ldots&0&0&\ldots&0\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\0&\ldots&1&0&\ldots&0\\\frac{\partial F_1}{\partial x_1}(x,y)&\ldots&\frac{\partial F_1}{\partial x_n}(x,y)&\frac{\partial F_1}{\partial y_1}&\ldots&\frac{\partial F_1}{\partial y_m}(x,y)\\\vdots&\ddots&\vdots&\vdots&\ddots&\vdots\\\frac{\partial F_m}{\partial x_1}(x,y)&\ldots&\frac{\partial F_m}{\partial x_n}(x,y)&\frac{\partial F_m}{\partial y_1}(x,y)&\ldots&\frac{\partial F_m}{\partial y_m}(x,y)\end{bmatrix}$

Due to the structure of this matrix, $J_G(x,y)=\det\frac{\partial F}{\partial y}(x,y),$ hence $J_G(x^0,y^0)\ne 0$ so we can apply the Inverse Function Theorem: there is an open neighbourhood $S$ of the point $(x^0,y^0),$ and open neighbourhood $W$ of $G(x^0,y^0)=(x^0,0)$ and the inverse function $G^{-1}:W\to S$ of the class $C^p$.

Since $S$ is open, there is an open neighbourhood $U$ of $x^0$ and an open neighbourhood $V$ of $y^0$ s. t. $U\times V\subset S$. Since $G^{-1}$ is continuous, the set $Y=G(U\times V)\subseteq W$ is open. Therefore, $G:U\times V\to Y$ is a $C^p$ -diffeomorphism.

It holds: $(x,y)=G^{-1}(G(x,y))=G^{-1}(x,F(x,y)), (x,y)\in U\times V.$

For  $(x,y)\in U\times V,$ which satisfy $F(x,y)=0\tag 1,$

it follows they are dependent through $(x,y)=G^{-1}(x,0).$ Therefore, we define a function $f:U\to V$ by $$f(x)=\pi_2\circ G^{-1}(x,0),$$ where $\pi_1(x,y)=x,\pi_2(x,y)=y.$ As $\pi_2$ and $G^{-1}$ are of the class $C^p,$ the function $f$ is, too, of the class $C^p$.

We now use the other composition $\begin{aligned}(x,0)&=G(G^{-1}(x,0))\\&=G(\pi_1\circ G^{-1}(x,0),\pi_2\circ G^{-1}(x,0))\\&=(\pi_1\circ G^{-1}(x,0), F(\pi_1\circ G^{-1}(x,0),\pi_2\circ G^{-1}(x,0)))\end{aligned}$

so that we conclude $x=\pi_1\circ G^{-1}(x,0),0=F(\pi_1\circ G^{-1}(x,0),\pi_2\circ G^{-1}(x,0)).$

From the definition of the function $f,$ it follows that $$0=F(x,f(x)),x\in U.\boxed{}$$

I have a question regarding $(1)$:

It says :

For $(x,y)\in U\times V$ which satisfy $F(x,y)=0,$... we define $f:U\to V$ by $f(x)=\pi_2\circ G^{-1}(x,0).$

I know that, since $Y=G(U\times V)$ is open and $G(x^0,y^0)=(x^0,F(x^0,y^0))=(x^0,0)\in Y,$ there certainly must be some other points of the form $(x,0)\in Y,$ where $x\in U,$ but how do we know $G^{-1}(x,0)$ isn't empty for an arbitrary $x\in U,$ i. e., that $f$ is well-defined $\forall x\in U$?

There is a mistake in the proof. You can consider $F(x,y)=y+x^2$ around $(0,0)$ to get an explicit (counter)example.

You can fix it by restricting $U$ to $U_{smaller}$ such that $U_{smaller}\times O \subseteq Y$ for some open $O$ containing $0$ and $U_{smaller}$ containing $x_0$. Then your $f$ is from $U_{smaller}$ to $V$ and is unique such.