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.

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 $\varepsilon >0$.
