For $S \subseteq \mathbb{R}^n$, let $f(S) = \{p \in \mathbb{R}^n\; |\;\text{d}(p,q) \le 1 \text{ for some }q \text{ in }S\}$.
In the thread
$\qquad$Given a convex set in a normed vector space, take a neighbourhood of it. Is still convex?
it was shown that if $S$ is convex, then so is $f(S)$.
It's easy to show that the converse is false.
But what about this question ...
Does there exist a bounded, nonempty set $S \subset \mathbb{R}^n$ such that the sets $$S, f(S), f(f(S)),...$$ are all non-convex?
In $\mathbb{R}^n$, we find that $f^k(S) = \{p \in \mathbb{R}^n \mid \exists q \in S \ d(p,q) \leq k\}$. This is because we can always pick a point on the line between $p$ and $q$ to witness the iterative containment.
In then follows that in $\mathbb{R}$ (so the one-dimensional case), for bounded $S$, some $f^k(S)$ will be an interval, and hence convex (just make $k$ bigger than the radius of $S$).
In higher dimensions, just consider a set with the two points $(0,...,0)$ and $(1,...,0)$. Assuming Euclidean distance, we find that $(0,k,0...,0)$ and $(1,k,0...,0)$ are in $f^k(S)$, but $(0.5,k,0,...,0)$ is not.