We can easily prove it by induction. The base case $n=4$ is obvious, suppose the result is true for $n=k$. The non-isolated subsets of $\{1,2,\ldots,k+1\}$ which does not contain $k+1$ is $(k-4)^2$. The non-isolated subsets which contain $k+1$ must be of the form $\{k+1,k,k-1,l,l-1\}$ ($2\le l\le k-2$) or $\{k+1,k,l,l-1,l-2\}$ ($3\le l\le k-2$), so there are $2k-7$ such subsets. In total, the number of non-isolated subsets is $(k-4)^2+2k-7=k^2-6k+9=((k+1)-4)^2$, as desired.