For the lower bound, consider the family \(\cgF\) of all \(k\) element subset of \([n]\) that contain \(1\text{.}\)
For the upper bound, let \(\cgF\) be a family of subsets of \([n]\) satisfying the two constraints. We show that \(|\cgF|\le\binom{n-1}{k-1}\text{.}\) To accomplish this, we consider a circle in the Euclidean plane with \(n\) points \(p_1\text{,}\) \(p_2,\dots,p_n\) equally spaced points around its circumference. Then there are \(n!\) different ways (one for each permutation \(\sigma\) of \([n]\)) to place the integers in \([n]\) at the points in \(\{p_1,p_2,\dots,p_n\}\) in one to one manner.
For each permutation \(\sigma\) of \([n]\text{,}\) let \(\cgF(\sigma)\) denote the subfamily of \(\cgF\) consisting of all sets \(S\) from \(\cgF\) whose elements occur in a consecutive block around the circle. Then let \(t=\sum_\sigma|\cgF(\sigma)|\text{.}\)
Our first claim is that \(t\le kn!\text{.}\) To prove this, let \(\sigma\) be a permutation and suppose that \(|\cgF(\sigma)|=s \ge 1\text{.}\) Then the union of the sets from \(\cgF(\sigma)\) is a set of points that form a consecutive block of points on the circle. Note that since \(n\ge 2k\text{,}\) this block does not encompass the entire circle. Accordingly there is a set \(S\) whose elements are the first \(k\) in a clockwise sense within this block. Since each other set in \(\cgF\) represents a clockwise shift of one of more positions, it follows immediately that \(|\cgF|\le k\text{.}\) Since there are \(n!\) permutations, the claim follows.
We now claim that for each set \(S\in\cgF\text{,}\) there are exactly \(n k!(n-k)!\) permutations \(\sigma\) for which \(S\in\cgF(\sigma)\text{.}\) Note that there are \(n\) positions around the circle and each can be used as the first point in a block of \(k\) consecutive positions in which the elements of \(S\) can be placed. Then there are \(k!\) ways to order the elements of \(S\) and \((n-k)!\) ways to order the remaining elements. This proves our claim.
To complete the proof of the theorem, we note that we have
\begin{equation*}
|\cgF| n k! (n-k)!\le t\le k n!,
\end{equation*}
and this implies that \(|\cgF|le\binom{n-1}{k-1}\text{.}\)