Consider a closed set $X^* \subset \mathbb{R}^n$. Let $Proj_{X^*}(x)$ denote the set of metric projections of $x \in \mathbb{R}^d$ to $X^*$: $$Proj_{X^*}(x) = \arg\min_{x^* \in X^*} d(x, x^*)$$ where $d(x, y) = \|x - y\|$.
Question: Given a neighborhood $S$ of $X^*$, is it always possible to find a neighborhood $S' \subset S$ of $X^*$ such that for all $y \in S'$, there is a $p \in Proj_{X^*}(y)$ such that the line segment connecting $y$ and $p$ is contained in $S'$?
Note 1: I want $S'$ to be a subset of $S$ (and so we cannot just take $S' = \mathbb{R}^n$).
Note 2: The line segment between $y$ and $p$ is the set $\{t y + (1-t) p : t \in [0,1]\}$.
My thoughts:
(1) My guess is the answer is yes, and in fact the following stronger statement is true: the line segment connecting $y$ and $p$ is contained in $S'$ for all $p \in Proj_{X^*}(y)$.
(2) If $X^*$ is compact, then we know the answer is yes because in this case there exists a $\tau > 0$ such that $S' := \{y : d(X^*, y) < \tau\} \subset S$. (Of course, we define $d(X^*, y) = \min_{x^* \in X^*} d(x, x^*)$.)
(3) If $X^*$ is not compact, I am not sure how to show this. I tried reducing it to the compact case, e.g., intersecting the set $X^*$ with larger and larger balls, or also cutting the set $X^*$ into disjoint pieces, but I'm unable to make it work.
(4) Another idea: We can also assume $S$ is of the form $S = \bigcup_{x^* \in X^*} B(x^*, r(x^*))$ with $r(x^*) > 0$, but I'm not able to use this formulation to prove the statement.