For any two subsets belonging to such a collection, the sets of the $k-1$ smallest elements must be different (or else they would be neighbours). There are $\binom{n-1}{k-1}$ ways to choose the $k-1$ smallest elements, since the number $n$ cannot be one of them. Therefore $f(n,k) \le \binom{n-1}{k-1}$, as desired.

For all $n \ge 2$ we have $f(n,1) = 1$, since any two 1-element sets are neighbours. So for all $n \ge 2$ and $k=1$ the statement holds. We will prove the general statement by induction by $n$, taking the base case to be $n=2$. We will assume that the statement holds for $n-1$ and consider the $n$-element set $S = \{1, 2, \dots, n\}$. Let $1 < k \le n-1$. The number of $k$-element subsets containing the number $n$ can be at most $f(n-1, k-1)$ in a neighbour-free collection, since removing the number $n$ from all of those subsets yields a neighbour-free collection of $k-1$-element subsets of $\{1, 2, \dots, n-1\}$. The number of $k$-element subsets not containing $n$ can be at most $f(n-1, k)$. Therefore $f(n,k) \le f(n-1, k-1) + f(n-1, k)$. By the induction assumption $f(n-1, k-1) \le \binom{n-2}{k-2}$ and $f(n-1, k) \le \binom{n-2}{k-1}$. Therefore by Pascal's rule $f(n,k) \le f(n-1, k-1) + f(n-1, k) \le \binom{n-2}{k-2} + \binom{n-2}{k-1} = \binom{n-1}{k-1}$, as desired.