Let be $S=\{a,b,c,d\}$, where $a,b,c,d\in\{1,2,\dots,n\}, a<b<c<d$. $b-a\ge1; c-b\ge 1; d-c\ge1\Longrightarrow d-a\ge3$. $a\in S$ and $a-1\notin S\Longrightarrow a+1\in S\Longrightarrow b=a+1$. $d\in S$ and $d+1\notin S\Longrightarrow d-1\in S\Longrightarrow c=d-1$. Hence: $S=\{a,a+1,d-1,d\}; d-a\ge 3$. The number of the sets $S$ is equal to the number of pairs $(a,d)$, where $a,d\in\{1,2,\dots,n\}, d-a\ge3$. Obviously, $n\ge 4$. For $a=1$ results $d\in\{4,5,\dots,n\}$, hence $n-3$ possible values of $d$. For $a=2$ results $d\in\{5,6,\dots,n\}$, hence $n-4$ possible values of $d$. ...... For $a=n-3$ results $d=n$, hence $1$ possible value of $d$. Results the number of possible sets $S$: $N_S=(n-3)+(n-4)+\dots+1=\dfrac{(n-3)(n-2)}{2}$.

The question is equivalent to pick two numbers from $\{1,2,..,n-1\}$ such that they have an absolute difference of at least two. See #12 for a generalization. The answer is $\binom{n-2}{2}$.

Given any set $S$, we can rearrange the elements in ascending order where $S=\{a_1,a_2,a_3,a_4\}, a_1<a_2<a_3<a_4$ and also we must have $\{a_1,a_2\}$ and $\{a_3,a_4\}$ are consecutive for the property b to hold. We claim that the number of neighbouring subsets is equivalent to choosing $2$ numbers, $\{a_2, a_3\}$ from $\{2,3,\ldots ,n-1\}$ where $a_2<a_3$ and $a_1=a_2-1, a_4=a_3+1$. It is easy to see that given any set $S$, we can arrange it into $\{a_1,a_2,a_3,a_4\}$ as desired. Also, observe that after rearranging any neighbouring subset in ascending order as described above, if we know $a_2,a_3$ then $a_1,a_4$ can be determined since by property b, this forces $a_1=a_2-1, a_4=a_3+1$ where $a_2,a_3 \in \{2,3, \ldots ,n-1\}$. Hence, the number of neighbouring subsets is $\boxed{\binom{n-2}{2}}$.

WLOG let the $4$ elements be $a, b, c, d$ such that $a<b<c<d$. Then if we pick any two elements $b, c$ we will know the other two elements $(a, d)$. Now because we can't let $b, c$ be the endpoints or else $a, d$ wouldn't be in the set $\{1, 2, \dots, n \}$. So we have to pick it out of $n-2$ elements which is $\binom{n-2}{2}$.

Let $S=\{a,b,c,d\}$ with $a<b<c<d$. By condition $b$, we have $b=a+1$ and $c=d-1$, so $S=\{b-1,b,c,c+1\}$. Since $b-1,c+1\in S$ we need $\{1,n\}\cap\{b,c\}=\emptyset$. All other ways work, there are $\boxed{\binom{n-2}2}$ such subsets.

Disclaimer: I can't use latex yet since I'm a new member. (This is just meant to be an outline of a solution that might be easier to understand for some people.) Let the set {1, 2, 3, ..., n} = N Using smaller examples like n=4, 5, 6, and 7, I noticed the solution would be (n-3)(n-2)½ but obviously I still had to prove why. Now I considered the ordered ascending set N and how I counted the number of neighbouring subsets. I chose the first two numbers (1, 2) and then counted the pairs of consecutive numbers from the remaining n-2 numbers, this would be the number of neighbouring subsets from N which contained (1,2). Then I repeated this process for (2, 3), (3, 4), ..., (n-1,n) which got the total number of neighbouring subsets. For the purpose of the solution, my 'algorithm' repeats for (n-2, n-1), and (n-1, n) even though they don't add any subsets. Consider this example: if n = 5, then N = {1, 2, 3, 4, 5}, and the neighbouring subsets of N are as follows: {1, 2, 3, 4} {1, 2, 4, 5} {2, 3, 4, 5} using the 'algorithm' I stated before. If you reverse this 'algorithm' (so start with (n-1, n) and then (n-2, n-1) and so on...), then it is clear you get a sum looking like 0+0+1+3+6+... which leads you to the solution.