Why is the Jacobian determinant continous in the proof of the Inverse function theorem. 
Can anyone explain to me why the first sentence in the proof is valid.
 A: Determinant is continuous on operators over $R^n$ (it is a polynomial of the coefficients with respect to the canonical, say, basis) and the jacobian matrix is also continuous by assumption. Composition preserves continuity. 
A: Alternatively to the answer of Oskar Linka, we can also see the continuity by looking at the Leibniz-formula for determinants.
First of all, note that $f \in C'$. (I guess $C'$ denotes the set of all maps whose partial derivatives exist in all variables and are continuous, i.e. all the maps $\frac{\partial f_i}{\partial x_j}$ exist and are continuous.) Now, using the Leibniz-formula, we have
$$J_f(b)=\sum_{\pi \in S_n} \text{sgn}(\pi)\prod_{i=1}^n \frac{\partial f_i}{\partial x_{\pi(i)}}(b)$$
for all $b \in S$. From this formula, we can directly read off the continuity of $J_f$ on $S$. It follows from the continuity of the maps $\frac{\partial f_i}{\partial x_j}$.
Now to the part about the open ball: if a map $h$ is continuous at some point $a$ and $h(a) \neq 0$, (e.g. $h=J_f$), there is always and open ball $B_{\varepsilon}(a)$ such that $h(x) \neq 0$ for all $x \in B_{\varepsilon}(a)$. Proof: Assume that this is not true. Then, for each $n \in \mathbb{N}$ there is an $x_n \in B_{\frac{1}{n}}(a)$ such that $h(x_n)=0$. Now, the sequence $(x_n)$ tends to $a$ for $n \to \infty$. However, $h(x_n)=0$ for all $x_n$. Therefore $h(x_n) \to 0$ for $n \to \infty$. But this is a contradiction: since $h$ is continuous at $a$, we should have $h(x_n) \to h(a) \neq 0$ instead. 
Note: $B_{\varepsilon}(a)$ denotes the open $\varepsilon$-ball around $a$ where it is assumed that $\varepsilon >0$.
