I'm trying to comprehend the solution of the following problem, from the course in elementary differential geometry that I'm taking.
Let $\gamma:(a, b) \rightarrow \mathbb{R}^{3}$ be a regular curve, and let $S$ be the surface parametrised by $$ \phi(u, v)=\gamma(u)+v b(u), \quad a<u<b,-\epsilon<v<\epsilon $$ where $b(u)$ is the binormal vector to $\gamma$ at time $u$. Show that there is $\epsilon>0$ so that $S$ is a regular surface, and prove that $\gamma$ is a geodesic in $S$.
Solution: Assume without loss of generality that $\gamma$ is parametrised by arc length, and let $T, n, b$ denote the Frenet frame at $\gamma(t)$. Since $\phi_{u}(t, 0)=\gamma^{\prime}(t)=T(t)$ and $\phi_{v}(t, 0)=b(t)$, the vector $$ \left(\phi_{u} \times \phi_{v}\right)(t, 0)=T(t) \times b(t)=-n(t) $$ is normal to $S$ at $\gamma(t)=\phi(t, 0),$ and hence $N(\phi(t, 0))=\pm n(t) .$ The geodesic curvature of $\gamma$ at $\gamma(t)=\phi(t, 0)$ is then given by $$ k_{g}(t)=\left\langle\gamma^{\prime \prime}(t),\left(N(\phi(t, 0)) \times \gamma^{\prime}(t)\right)\right\rangle=\langle k(t) n(t), \pm n(t) \times T(t)\rangle=0 $$ since $n \times T=-b$ is orthogonal to $n$.
I understand the written lines above, but I don't understand why this implies that there exists an $\epsilon$ such that $S$ is a regular surface. Could anyone help me with this? (If I'm indeed not missing something, how can I prove the existence of such $\epsilon$?)