The second point is easy: take $f_n \equiv 0$.

If the feasible region is bounded, my conclusion fails. The main argument was that we can look at rays starting in $0$.

Your example does not have a minimizer. If the problem has a minimizer, then $0$ is a minimizer. Otherwise, the infimal value is $-\infty$.

Is $S$ closed in $L^2$?

I don't get your point concerning orthogonality. If $(v_1,\ldots,v_k,y)$ is orthogonal, then $x$ is orthogonal to $y$. This implies $\beta = 0$. Thus, $\|x -\beta\,y\|/\|x\| = 1$?

I do not believe that your first condition is necessary: If $f$ is the indicator function of $\{0\}$, then the proximal operator is $x \mapsto 0$ and this is not invertible. Second, the domain of a proximal operator is always $H$, i.e., it is trivially convex.

Yes, of course.

Do you want a bound using only $\alpha$ and $\beta$?