Existence of a neighborhood whose points have secents perpendicular to a given arc The following seems like an "obvious" geometric fact, but I really can't figure out how to begin a proof of it... I can tell it will require continuity of $\gamma^\prime$, but that's all I can really say at this point.  Any ideas or hits are most appreciated.
Let $\gamma:[-\ell,\ell]\to\mathbb{R}^n$ be a simple $C^2$ curve parameterized with respect to arclength, and let $\delta>0$.  Show there exists some $d>0$ such that for all $c\in B_d(\gamma(0))$ that there exists $s_0\in[-\delta,\delta]$ such that $c-\gamma(s_0)\perp\gamma^\prime(s_0)$.
This statement can be better understood in the context of this diagram:
 A: You are looking for the shortest connection from $c$ to $\gamma$. If you already have a straight line from $c$ to $\gamma$ which realizes this, then the function 
$$t\mapsto ||c-\gamma(t)||^2 $$ 
will attain a minimum in the point $t_0$ where this line meets $\gamma$, so 
$$\frac{d }{dt} ||c-\gamma(t)||^2 = 2\langle c-\gamma(t_0), \gamma^\prime(t_0)\rangle = 0 $$
which just states the orthogonality relation.
There are several possible approaches to find such a line. One is to look at  a continuous choice $n(t)$ of a unit normal vector along $\gamma$, which is of class $C^1$ (it's a locally unique choice of an orthonormal vector to $\gamma^\prime$). Then look at 
$$(s,t) \mapsto\phi(s,t) := \gamma(t) + s n(t)$$
It's not difficult to see that $d\phi(0, t) = id$ is a linear isomorphism for each $t$, so by the inverse function theorem the map $\phi$ is a diffeomorphism in a neighbourhood of $(0,t)$ for each $t$. In particular the map is onto in a $\delta$- neighbourhood of, say $[t_0-\alpha, t_0+\alpha]$ for some $\alpha >0$ and covers a neighbourhood of the image of the curve near $t_0$. This is the neighbourhood you are looking for (each point $c$ in this neighbourhood is by construction connected to $\gamma$ by a line orthogonal to $\gamma$.
