Let $M$ be a smooth Riemannian manifold, and let $S \subseteq M$ be a compact submanifold.
Define $U=\{ p \in M \, | \, \text{there exist a unique closest point to } p \text{ in } S\}$. Is $U$ open in $M$?
Does the answer change if we assume that $M$ is complete or geodesically convex?
Nope. Here's a counterexample. Let $M=\mathbb R^2$, and let $S$ be the ellipse defined by $x^2/4 + y^2 = 1$. For a point $p=(a,0)$ on the positive $x$-axis, a nice calculus exercise shows the following: As long as $0< a< 3/2$, there are two points on the ellipse closest to $p$, namely those points whose $x$-coordinate is $4a/3$. But for $3/2 \le a < 2$, the unique closest point to $p$ is $(2,0)$. In particular, $p=(3/2,0)$ has no neighborhood contained in $U$.