[proofplan] We first identify the neighborhood complex of the Kneser graph with the simplicial complex of finite families of $k$-subsets whose union has size at most $n-k$. We then pass from this complex to the poset of subsets of $[n]$ whose cardinalities lie between $k$ and $n-k$, using the union map and the Quillen fiber criterion for finite posets. Finally, we use the rank-selection connectivity theorem for the Boolean lattice to identify the connectivity of the resulting order complex. This gives the required $(n-2k-1)$-connectivity. [/proofplan] [step:Identify the neighborhood complex with a union-size complex] Let $\binom{[n]}{k}$ denote the set of all $k$-element subsets of $[n]$. Define a simplicial complex $\mathcal{M}$ with vertex set $\binom{[n]}{k}$ by declaring a finite subset $\mathcal{A} \subset \binom{[n]}{k}$ to be a face when \begin{align*} \left|\bigcup_{A \in \mathcal{A}} A\right| \leq n-k. \end{align*} We claim that $N(KG(n,k))=\mathcal{M}$ as simplicial complexes. Indeed, a finite family $\mathcal{A} \subset \binom{[n]}{k}$ is a face of the neighborhood complex $N(KG(n,k))$ exactly when the vertices in $\mathcal{A}$ have a common neighbor in $KG(n,k)$. By the definition of the Kneser graph, this means that there exists a $k$-element subset $B \subset [n]$ such that $B \cap A = \varnothing$ for every $A \in \mathcal{A}$. Such a set $B$ exists exactly when the complement \begin{align*} [n] \setminus \bigcup_{A \in \mathcal{A}} A \end{align*} has cardinality at least $k$, which is equivalent to \begin{align*} \left|\bigcup_{A \in \mathcal{A}} A\right| \leq n-k. \end{align*} Thus the faces of $N(KG(n,k))$ and $\mathcal{M}$ coincide. [/step] [step:Replace the union-size complex by a rank-selected Boolean poset] Let $P$ denote the poset, ordered by inclusion, whose elements are the subsets $S \subset [n]$ satisfying \begin{align*} k \leq |S| \leq n-k. \end{align*} Let $\operatorname{Face}(\mathcal{M})$ denote the face poset of $\mathcal{M}$, ordered by inclusion; its elements are all faces of $\mathcal{M}$, including $\varnothing$. Let $\operatorname{Sd}(\mathcal{M})$ denote the barycentric subdivision of $\mathcal{M}$, whose vertices are the nonempty faces of $\mathcal{M}$ and whose faces are chains of nonempty faces ordered by inclusion. Define the order-preserving map $u: \operatorname{Face}(\mathcal{M}) \setminus \{\varnothing\} \to P$ by \begin{align*} u(\mathcal{A}) := \bigcup_{A \in \mathcal{A}} A. \end{align*} This map is well-defined because every nonempty face $\mathcal{A}$ contains at least one $k$-element subset, so $|u(\mathcal{A})| \geq k$, and the defining condition for $\mathcal{M}$ gives $|u(\mathcal{A})| \leq n-k$. For each $S \in P$, let $P_{\leq S}$ denote the subposet of all $T \in P$ with $T \subset S$. The lower fiber $u^{-1}(P_{\leq S})$ is the face poset of the full simplex on the vertex set $\binom{S}{k}$, with the empty face removed. Since $|S| \geq k$, this vertex set is nonempty, and the corresponding order complex is the barycentric subdivision of a nonempty simplex. Hence every such lower fiber has contractible order complex. By the finite-poset form of Quillen's Theorem A, also called the Quillen fiber criterion for finite posets, the induced simplicial map $\operatorname{Sd}(\mathcal{M}) \to \Delta(P)$ is a homotopy equivalence, where $\Delta(P)$ denotes the order complex of $P$. Since barycentric subdivision preserves homotopy type, $\mathcal{M}$ is homotopy equivalent to $\Delta(P)$. [guided] The earlier union-size description of $\mathcal{M}$ suggests recording the union of a face. We make that precise by introducing the poset $P$ of all subsets $S \subset [n]$ whose sizes lie in the range \begin{align*} k \leq |S| \leq n-k. \end{align*} The lower bound $k$ appears because every nonempty face of $\mathcal{M}$ contains at least one $k$-subset. The upper bound $n-k$ is exactly the defining union-size condition for $\mathcal{M}$. Let $\operatorname{Face}(\mathcal{M})$ denote the face poset of $\mathcal{M}$, ordered by inclusion; its elements are all faces of $\mathcal{M}$, including $\varnothing$. Let $\operatorname{Sd}(\mathcal{M})$ be the barycentric subdivision of $\mathcal{M}$. Its vertices are the nonempty faces of $\mathcal{M}$, and its simplices are chains of such faces. Define the union map $u: \operatorname{Face}(\mathcal{M}) \setminus \{\varnothing\} \to P$ by \begin{align*} u(\mathcal{A}) := \bigcup_{A \in \mathcal{A}} A. \end{align*} This is an order-preserving map: if $\mathcal{A} \subset \mathcal{B}$ as faces, then $u(\mathcal{A}) \subset u(\mathcal{B})$. It is also well-defined. Since $\mathcal{A}$ is nonempty, it contains some $k$-element subset $A$, so $|u(\mathcal{A})| \geq |A|=k$. Since $\mathcal{A}$ is a face of $\mathcal{M}$, we also have \begin{align*} |u(\mathcal{A})| = \left|\bigcup_{A \in \mathcal{A}} A\right| \leq n-k. \end{align*} Therefore $u(\mathcal{A}) \in P$. We now verify the hypothesis needed for the Quillen fiber criterion for finite posets. Fix $S \in P$, and let $P_{\leq S}$ denote the subposet of all $T \in P$ satisfying $T \subset S$. The lower fiber $u^{-1}(P_{\leq S})$ consists exactly of those nonempty faces $\mathcal{A}$ of $\mathcal{M}$ whose union is contained in $S$. This is the same as saying that every vertex $A \in \mathcal{A}$ is a $k$-element subset of $S$. Hence the lower fiber is the face poset, with the empty face removed, of the full simplex on the vertex set $\binom{S}{k}$. Because $|S| \geq k$, the set $\binom{S}{k}$ has at least one element, so this simplex is nonempty and contractible. Its barycentric subdivision, which is the order complex of the lower fiber, is also contractible. The finite-poset form of Quillen's Theorem A, also called the Quillen fiber criterion for finite posets, now applies: an order-preserving map whose lower fibers have contractible order complexes induces a homotopy equivalence on order complexes. Thus the induced map $\operatorname{Sd}(\mathcal{M}) \to \Delta(P)$ is a homotopy equivalence. Finally, barycentric subdivision does not change homotopy type, so $\mathcal{M}$ is homotopy equivalent to the order complex $\Delta(P)$. [/guided] [/step] [step:Apply rank-selection connectivity for the Boolean lattice] Let $B_n$ denote the Boolean lattice of all subsets of $[n]$, ordered by inclusion, and let $\rho: B_n \to \{0,1,\dots,n\}$ be the rank function defined by $\rho(S):=|S|$. The poset $P$ is the rank-selected subposet of $B_n$ with ranks $k,k+1,\dots,n-k$. Set \begin{align*} r := n-2k. \end{align*} The rank-selection connectivity theorem for the Boolean lattice, equivalently the rank-selection connectivity theorem for the Cohen-Macaulay Boolean lattice, says that, for integers $0 \leq a \leq b \leq n$, the order complex of the subposet of $B_n$ consisting of ranks $a,a+1,\dots,b$ is $(b-a-1)$-connected. We verify its numerical hypotheses for $a:=k$ and $b:=n-k$: since $k$ is a positive integer and $n \geq 2k$, we have $0 \leq k \leq n-k \leq n$. Applying the theorem with these values, we obtain that $\Delta(P)$ is \begin{align*} (n-k-k-1) \end{align*} -connected. Since $n-k-k-1=n-2k-1=r-1$, $\Delta(P)$ is $(n-2k-1)$-connected. [/step] [step:Transfer the conclusion back to the Kneser neighborhood complex] From the first step, $N(KG(n,k))=\mathcal{M}$ as simplicial complexes. From the union-map comparison, $\mathcal{M}$ is homotopy equivalent to $\Delta(P)$, and the previous step proves that $\Delta(P)$ is $(n-2k-1)$-connected. Connectivity is invariant under homotopy equivalence, so $\mathcal{M}$ is $(n-2k-1)$-connected. Therefore $N(KG(n,k))$ is $(n-2k-1)$-connected, as required. [/step]