Doubt in proof of special case of implicit function theorem

I've been studying the implicit and inverse functions theorems and I've started with one special case of the implicit function theorem. The book I'm reading states the theorem as follows:

Let $f : U\to\mathbb{R}$ be a function of class $C^k$ with $k\geq 1$ defined on some open set $U\subset\mathbb{R}^n\times\mathbb{R}$. If $p=(x_0,y_0)\in U$ is such that $f(p)=c$ and $D_{n+1}f(p)\neq0$, then there exists a ball $B(x_0;\delta)\subset \mathbb{R}^n$ and one interval $J=(y_0-\epsilon,y_0+\epsilon)$ such that $f^{-1}(c)\cap (B\times J)$ is the graph of a function $\xi : B(x_0;\delta)\to J$ of class $C^k$ and for all $x\in B(x_0;\delta)$ we have $D_i\xi(x)=-D_if(x,\xi(x))/D_{n+1}f(x,\xi(x))$.

Now, the proof starts as follows:

Consider that $D_{n+1}f(x_0,y_0)>0$, then since $D_{n+1}f$ is continuous, there exists $\delta>0$ and $\epsilon>0$ such that if $B$ is the open ball with center $x_0$ and radius $\delta$ and if $J=(y_0-\epsilon,y_0+\epsilon)$ then $B\times \bar{J}\subset U$ and $D_{n+1}f(x,y)>0$ for all $(x,y)\in B\times \bar{J}$.

My doubt is exactly there. First: why can we assure that $B\times\bar{J}\subset U$ based on continuity? I know that if some continuous function is positive in some point, then there's a neighbourhood of the point where it is still continuous, but why can we take the neighbourhood like that? That closure confused me a little.

The only thing I could think of was the following: if we endow $\mathbb{R}^n\times\mathbb{R}$ with the product topology and endow $\mathbb{R}^n,\mathbb{R}$ with the metric topology, then a set of $\mathbb{R}^n\times\mathbb{R}$ is open if and only if for every point of the set there's one product $U_1\times U_2$ of open sets $U_1\subset\mathbb{R}^n,U_2\subset\mathbb{R}$. Now, since $D_{n+1}f$ is continuous and $D_{n+1}f(p)>0$, then there's a neighbourhood $N(p)\subset U$ such that for every $q\in N(p)$ we have $D_{n+1}f(q)>0$. Since it is a neighbourhood of $p$, there must be inside of it an open set containing $p$, now we can take a basic one, because the definition of the open sets. But why can we assure that the product of the ball with the closure of the interval is still in $N(p)$?

If $B\times J \subset U$ but $B\times \overline J$ isn't in $U$, then just replace $J$ by $J' = (y_0 - \epsilon/2,y_0 + \epsilon/2)$.
• But I've said: "a set of $\mathbb{R}^n\times\mathbb{R}$ is open if and only if for every point of the set there's one product $U_1\times U_2$ of open sets $U_1\subset\mathbb{R}^n, U_2\subset \mathbb{R}$", isn't that the way we define a topology generated by a basis? At every point of the open set there's a basic open set containing the point still inside the set? Now I've got the thing with the closure! Thanks very much @BaronVT! – user1620696 Oct 8 '13 at 19:23