Here is a solution for $n=2d^2-2$ with $|S|=4d-3.$ It is just Gerry Myerson's scheme with a very slight adjustment, so the same asymptotics.
$$0,1,\cdots d-1$$ then $$2d-1,3d-1,4d-1,\cdots ,(2d-2)d-1$$ then $$n-d+1,n-d+2,\cdots,n.$$
Note that $n-d+1=(2d-1)d-1.$
Here are some of Rob Prats's reported values (which I have no reason to question) with $n=2d^2-2$ in bold.
$|S|\ \ \ n$
$5\ \ \ \ \ \ 5,\mathbf{6}$
$9\ \ \ \ \ \ 14,15,\mathbf{16}$
$13\ \ \ \ \ \ 28,29,\mathbf{30},32$
$14\ \ \ \ \ \ 31,33,34,35,36$
$17\ \ \ \ \ \ 47,\mathbf{48},49,50$ and perhaps a few larger $n?$
It would look nicer if $n=31$ required $|S|=14,$ this is the only case reported with $|S|=m(n) \lt m(n-1).$
I would be curious to see the optimal sets $|S|.$