Use the Inverse Function Theorem to prove the following case of the Implicit Function Theorem.
If $f$ is a real-valued $C^1$ function on an open subset $U\subseteq \mathbb{R}^2$ with $f(a,b) = 0$, $\frac{\partial f}{\partial y}(a,b)\neq 0$ for some $(a,b)\in\mathbb{R}^2,$ show that there are open intervals $a\in I,$ $b\in J$ and a unique $C^1$ function $\phi:I\to J$ such that $f(x,\phi(x)) = 0$ on $I.$
I am following the standard proof but am hung up on a few things. First let $F(x,y) = (x,f(x,y))$ for $(x,y)\in\mathbb{R}^2,$ then $DF(a,b)$ is invertible, so by the Inverse Function Theorem there are open neighborhoods $U$ of $(a,b)$ and $V$ of $F(a,b) = (a,0)$ such that $F:U\to V$ has a $C^1$ inverse. Let $G_1,G_2:V\to\mathbb{R}$ be the coordinate functions of $F^{-1}$. If $(x,0)\in V$, then $$(x,0)= F(G_1(x,0),G_2(x,0)) = (G_1(x,0),f(G_1(x,0),G_2(x,0)))$$ implies $G_1(x,0) = x$ and $f(x,G_2(x,0)) = 0$.
Then I think we should take $I = \left\{x\in\mathbb{R}:(0,x)\in V\right\}$ and $\phi(x) = G_2(x,0),$ but I don't understand why $I$ is open.
Any clarification or especially a simpler proof would be greatly appreciated!
The function $G: \mathbb R\rightarrow \mathbb R^2$ given by $G(x)=(0, x)$ is continuous. Since $V$ is open, $I=G^{-1}[V]$ is open.