If $M$ is a $m$-dimensional differentiable manifold and $\omega$ is a continuous $m$-form on $M$ such that $\omega(x) \neq 0$ for every $x \in M$, then $M$ is orientable.
The author takes $A$ as the set of parameterizations $\varphi: U_0 \to U$ such that, for each point $x = \varphi(u)$, we have $\omega(x) = a(u)du_1 \wedge \cdots \wedge du_m$ with $a(u) > 0$. Assuming $U_0$ connected, we can see that $a > 0$ for every $u \in U_0$.
Here is my doubt: the author says it implies that $A$ is an atlas.
I cannot see why the parameterizations of $A$ covers $M$.
If $p \in M$ then let $\phi' : U_0 \to U \ni p$ be a chart containing $p$. Since $\omega$ is nowhere zero and a top-degree form there is a nonzero function $a : U \to \mathbb{R}$ such that $\omega = a(u)du_1 \wedge \cdots \wedge du_m$. This function $a$, by continuity, is either always positive or always negative. If it is always positive we are done. If it is always negative, then switch coordinates $u_1$ and $u_2$, say, so that after this switch, the new volume form (call it $dv_1 \wedge \cdots \wedge dv_m$) is of the opposite sign as $du_1 \wedge \cdots \wedge du_m$. The minus sign can be absorbed into the function $a$, so now $$ (-a(v))dv_1 \wedge \cdots \wedge dv_m = a(u) du_1 \wedge \cdots \wedge du_m = \omega. $$