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.
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.