I am reading Spivak Calculus on Manifold Chapter 5 and it has the following argument:
If $M$ is an oriented $(n-1)$-dimensional manifold in $R^n$, a substitute for outward unit normal vectors can be defined, even though M is not necessarily the boundary of an $n$-dimensional manifold. If $[v_1,...,v_{n-1}] = \mu_x$ (which is the orientation of $M$ at $x$), we choose $n(x)$ in $R^n_x$ so that $n(x)$ is a unit vector perpendicular to $M_x$ and $[n(x),v_1,...,v_{n-1}]$ is the usual orientation of $R^n_x$. The vectors $n(x)$ vary continuously on $M$, in an obvious sense.
I understand this approach of defining outward unit normal vectors. But why $n(x)$ vary continuously? Or why $n(x)$ is a smooth vector field?
$n(x)$ is uniquely determined up to sign. Now think about the determinant of the matrix with columns $n(x)$, $v_1(x), v_2(x), \dots, v_{n-1}(x)$. The sign on $n(x)$ is chosen to make the determinant positive. Remember that determinant is a continuous (smooth) function of the $n$ column vectors. If $v_i(x)$ vary continuously (smoothly) with $x$ (work in a coordinate chart to make this condition hold), then $n(x)$ must likewise vary continuously to keep the sign of the determinant positive.