Suppose $D \subset \mathbb{R}^n$ is a smoothly bounded domain given by a smooth defining function $r$. That is $$ D = \{r < 0\}, \quad \partial D = \{r = 0\}, \quad \text{ and } \nabla r|_{\partial D} \neq 0 $$
Given a boundary point $p \in \partial D$, why is it that for $\epsilon > 0$ sufficiently small, the sphere of radius $\epsilon$ centered at $p$, $$ S(p, \epsilon) = \{z: |z - p| = \epsilon\} $$ is tangent to a level set of $r$ at exactly one point in $D$?
I believe that one uses that in $D$ near $\partial D$ that $\nabla r$ does not vanish. Thus, the level sets $r^{-1}(c)$ are smooth hypersurfaces for $c$ with $|c|$ small.
Conceptually, as soon as $\epsilon$ is small enough that the curvature of the $\epsilon$-spehere is higher than that of the level hypersurfaces near $p$, the result will hold. Technically, you can argue as follows:
There is a smooth local diffeo $\Psi$ that transforms function $r$ into a linear function (usually taken to be $f(x_1,..., x_n)=x_1$. This is a special case of the vector field rectification theorem, using the fact that $\nabla r$ is non-zero. We can assume $\Psi(p)=\vec{0}$.
The image of any small enough ball around $p$ under a smooth map surjective at $p$ is strictly convex. This is Remark 4, p.161 here.
Now after applying $\Psi$ we are looking for points where $\Psi(S(p, \epsilon))$ is tangent to the level of $x_1$ with $x_1<0$. But $\Psi(S(p, \epsilon))$ is the boundary of compact convex set $\Psi(B(p, \epsilon))$, with $\vec{0}$ in the interior. There is precisely one such point, the one where $x_1$ attains its minimum on $\Psi(B(p, \epsilon))$ (convex set lies on one side of each tangent plane, so each tangency is a min or a max, and for strictly convex sets there is at most one of each).
You could also argue directly. Assume for simplicity that $\nabla r|_p$ is unit size. Suppose there are two points of tangency, with radius vectors of the sphere positively proportional to the gradient at each. Then denoting the angle on the sphere between them by $\theta$, we know that the distance between them is $d =2\sin( \theta/2) \epsilon$, and for small enough $\epsilon$ the vector $\nabla r$ can not rotate by that much over such small distance: the minimal distance between a unit vector and a vector at angle $\theta$ to it is $\sin (\theta)>\sin(\theta/2)$, which is greater than how much $\nabla r$ can change over a distance of $2\sin( \theta/2) \epsilon$ as soon as $1/(2\epsilon)$ is greater than the operator norm of the Hessian of $r$ near $p$. This again shows that there is at most one tangency of the small sphere with positive proportionality between $\nabla r$ and the radius vector. Similarly, there is at most one tangency with negative proportionality. Since there are definitely two tangencies - one where $r$ attains the maximum and one where it attains the minimum,- there is only one inside $D$ (the minumum one).
Some details:
Lemma: Consider a smooth $F:\mathbb{R}^n \to \mathbb{R}^m$ and $a, b \in \mathbb{R}^n$. Suppose that operator norm of $DF$ is less than $k$ everywhere on the segment $ab$. Then $|F(a)-F(b)| < k |a-b|$.
Proof: $$|F(a)-F(b)|=|\int_0^1 DF(ta+(1-t)b)(\frac{b-a}{|b-a|} |b-a|) dt| \leq \int_0^1 |F(ta+(1-t)b)(\frac{b-a}{|b-a|})| |b-a| dt < \int_0^1 k |b-a| dt = k |a-b|.$$
In our case $F(x)=\nabla r(x)$ so $DF$ is the Hessian at $x$. Sufficiently near $p$ the operator norm of the Hessian is bounded by twice the operator norm of the Hessian at $p$ (by continuity of operator norm). So it's bounded by some constant $M$. Now if $\frac{1}{2 \epsilon}>M$ and $\epsilon$ is small enough that the norm bounds hold on closed ball of radius $\epsilon$ around $p$, then for any two points $a$ and $b$ on the sphere (of radius $\epsilon$ centered at $p$) at distance $d$ from each other, by the lemma we have the distance from $\nabla r(a)$ and $\nabla r(b)$ is at most $d \cdot M < d/ (2 \epsilon)$, so if $d=2\sin( \theta/2) \epsilon$, the distance $|\nabla r(a)-\nabla r(b)| < \frac{2\sin( \theta/2) \epsilon }{(2 \epsilon)} < \sin(\theta)$, so the angle between them is less than $\theta$.