Projection of $\lambda y$ onto a polytope is fixed for sufficiently large $\lambda$? Le $y\in \mathbb R^n\setminus\{0_n\}$.
Let $X\subset \mathbb R^n$ be a compact polytope (intersection of finite half-spaces).
For a sufficiently large $\lambda\in \mathbb R_+$, I have the impression that the projection of $\lambda y$ onto $X$ is fixed. Is it correct?
More precisely, does the following statement hold true?
$$\exists (x_0, \lambda_0) \in X\times \mathbb R_+: \hspace{0.3cm} \text{proj}_X(\lambda y)=x_0, \,\, \forall \lambda\geq \lambda_0.$$

Here I explain how I came up with the assumptions.
First, $y\neq 0_n$ is a "non-trivial" condition, otherwise, $\lambda y\equiv 0_n$, and the statement is obvious.
Second, $X$ is closed since the projection of a point onto a set always belongs to the closure of that set.
Finally, $X$ should be bounded and non-smooth, since otherwise, the counter examples (in which $\text{proj}_X(\lambda y)$ is not fixed for all large $\lambda$) can be found in the following figure.
Note. Feel free to impose extra conditions if needed since the conditions listed here are just necessary conditions.

 A: Here is one unsatisfactory proof by induction on dimension.
There is no effective change in generality if we consider the projection of the half line $\{b+ty\}_{t \ge 0}$ and it will simplify the inductive step.
Here is the general idea behind the proof. We want to show that for a compact polytope that for any $b,y$ there is some $T$ such that for $t\ge T$ the projection of $b+ty$ onto $X$ is constant.
It is clear in one dimension, so we will use induction.
The compact polytope is defined by the intersection of half spaces and these define a finite number of faces $F_k$ of $X$, and for any point $x \notin X$, $\operatorname{proj}_X x$ lies in one of these faces. Furthermore, if $\operatorname{proj}_X x \in F_k$ then we must have $\operatorname{proj}_X x = \operatorname{proj}_{F_k} x$. If we can show that for each $k$ that there is some $T_k$ such that $\operatorname{proj}_{F_k} (b+ty)$ is constant (say $f_k$) for $t \ge T_k$, then for sufficiently large $t$, we have $\operatorname{proj}_{X} (b+ty) \in \{f_k\}_k$. Since $\operatorname{proj}_{X}$ is continuous, this means that, by connectedness, that $\operatorname{proj}_{X} (b+ty)$ is constant. To show that the projection onto $F_k$ is eventually constant, we project $b+ty$ onto the hyperplane containing $F_k$ which reduces the dimension of the problem by one and we can use induction.
It relies on the result that for a closed convex set $C$,
$\operatorname{proj}_C x = \operatorname{proj}_C 
 (\operatorname{proj}_{\operatorname{aff} C} x)$. That is, you can compute the projection by projecting onto the affine hull of $C$ first and then projecting onto $C$. This is the step that can reduce dimension.
Another relevant fact is that for a compact set, the map $x \mapsto \operatorname{proj}_C x$ is continuous.
By dimension, I mean the affine dimension of the set $X \cup \{b+ty\}_{t \ge 0}$.
The result is clear in one dimension since everything can be identified with $\mathbb{R}$, and $X$ is identified with a compact interval.
Assume the dimension is $n>1$ and that the result is true in smaller dimensions.
Suppose $X = \cap_k H_k$, where $H_k = \{ x \mid h_k^T x \le \beta_k \}$, for $k=1,...,m$.
Let $F_k = \{ x \in X \mid h_k^T x = \beta_k \}$, and let $\Pi_k$ be the orthogonal projection onto the hyperplane $P_k=\{x \mid h_k^T x = \beta_k \}$.
It is straightforward to show that $\Pi_k$ is affine.
Note that for a sufficiently large $t$ (so that the point is outside $X$) we see that $\operatorname{proj}_X (b+ty)$ lies in at least one $F_k$, and since $F_k \subset X$, we see that $\operatorname{proj}_X (b+ty)= \operatorname{proj}_{F_k} (b+ty)$.
Hence if we can show that for sufficiently large $t$ that $\operatorname{proj}_{F_k} (b+ty)$ is constant, we can use continuity of $ \operatorname{proj}_X$ to conclude the desired result.
Pick any $k$, then if $\Pi_k (b+ty)$ is constant the result is clear. If not, note that for any $x$, $\operatorname{proj}_{F_k} x = \operatorname{proj}_{F_k} 
 (\operatorname{proj}_{\operatorname{aff} F_k} x) = \operatorname{proj}_{F_k} \Pi_k x$, and since $F_k$ and $\Pi_k(b+ty)$ lie in $P_k$ (which has dimension $n-1$), we see that the result is true.
