Let $A$ be a compact subset of $\mathbb{R^n}$. Suppose that $A \cap V_c$ is contained in an open set $U$ in $V_c$. Then for any suitably small interval $I$ about $c$ in $\mathbb{R}$, $A \cap V_I$ is contained in $I \times U$.
$V_c$ is the ''vertical slice'' $\{c\} \times \mathbb{R^{n-1}}$ and $V_I$ is simply the set $I \times \mathbb{R^{n-1}}$, where $I$ is an interval such that $I \subset \mathbb{R}$
Proof : If not, there would exist a sequence of points $(x_j,c_j)$ in A such that $c_j \to c$ and $x_j \notin U$. Replace this sequence by a convergent one to get a contradiction.Q.E.D
I have a problem to understand well this proof. Someone could help me step by step to decrypt it?
Suppose $A\cap V_I$ is not contained in $I\times U$, for any interval $I$. Let $I_j = (c-\frac{1}{j},c+\frac{1}{j})$. Then there is $(c_j,x_j)\in A\cap V_{I_j}$ but $(c_j,x_j)\notin I_j\times U$. (I switched the roles of the $x$ and the $c$, I think it fits better with the established notation.)
Well, from the definition of $V_{I_j}$ we must have $c_j\in I_j$. Therefore if $(c_j,x_j)\notin I_j\times U$ it follows that $x_j\notin U$. As $j\to\infty$, since the $I_j$-s shrink to $c$ we see that $c_j\to c$.
Where do the $x_j$-s go? If we project $A$ to $\mathbb{R}^{n-1}$ (call the projection $\pi(A)$), then $x_j\in \pi(A)$ for all $j$. Since $A$ is compact the projection $\pi(A)$ is compact. Therefore the $x_j$ have a convergent subsequence, and the limit is in $\pi(A)$. Replacing $(c_j,x_j)$ by this convergent subsequence which we also call $(c_j,x_j)$, we have $$ (c_j,x_j)\to (c,x) \in \{c\}\times\pi(A) = A\cap V_c. $$ But wait! Each $x_j\notin U$, i.e. $x_j\in \mathbb{R}^{n-1}\setminus U$. $\mathbb{R}^{n-1}\setminus U$ is a closed set, so it contains its limit points. Since the $x_j\to x$, $x$ is a limit point, so $x\in\mathbb{R}^{n-1}\setminus U$, i.e. $x\notin U$. But $A\cap V_c \subset \{c\}\times U$, so this is a contradiction.
(There is some abuse of notation, which I guess I inherited from G&P: $U$ can be taken to mean the open set in $V_c = \{c\}\times\mathbb{R}^{n-1}$ containing $A\cap V_c$, or it can mean the set $U$ in $\{c\}\times U$, $U$ open in $\mathbb{R}^{n-1}$, such that $A\cap V_c \subset \{c\}\times U$. I'm using the latter.)