Yes.
Claim 1: $X(k,\infty)$ is contractible.
Proof: There is a functor $X(k,\infty)\times X(k,\infty)\to X(k,\infty)$ given by laying two ordered sets end to end. It admits a natural map from each of the two projection functors, and this implies that the resulting map of spaces (realizations of nerves) is homotopic to each of the two projections. But when a nonempty space $X$ is such that the two projections $X\times X\to X$ are homotopic then $X$ is contractible.
Claim 2: The relative homology $H_j(X(k_1,k_2+1),X(k_1,k_2))$ is trivial for $j\ne k_2-k_1+1$.
Proof: Let $P(k_1,k_2+1)$ be the poset of subsets of $\lbrace 0,\dots ,k_2+1\rbrace $ of cardinality at least $k_1+1$, and let $bP(k_1,k_2+1)$ be the subposet of proper subsets. There is a functor $P(k_1,k_2+1)\to X(k_1,k_2+1)$ inducing an isomorphism of quotients of nerves $P(k_1,k_2+1)/bP(k_1,k_2+1)\to X(k_1,k_2+1)/X(k_1,k_2)$. $P(k_1,k_2+1)$ is contractible, while $bP(k_1,k_2+1)$, being isomorphic to the $(k_2-k_1)$-skeleton of a $(k_2+1)$-simplex, has its reduced homology all in dimension $k_2-k_1$.
Claim 2 implies by induction on $m$ that the relative homology $H_j(X(k_1,k_2+m),X(k_1,k_2))$ is trivial for $j\le k_2-k_1$. Thus $H_j(X(k_1,\infty),X(k_1,k_2))$ is trivial for $j\le k_2-k_1$. But this is isomorphic to the (reduced) homology $\tilde H_{j-1}X(k_1,k_2)$, by Claim 1.
It follows that $X(k_1,k_2)$ has no homology below dimension $k_2-k_1$. Since you say you have checked that it is also $1$-connected (if $k_2-k_1\ge 2$), this makes it $(k_2-k_1-1)$-connected.
EDIT More simply: The inclusion map $X(k_1,k_2-1)\to X(k_1,k_2)$ is homotopic to a constant because it admits a natural map from a constant map. Therefore the cofiber $X(k_1,k_2)/X(k_1,k_2-1)$ is homotopy equivalent to $\Sigma X(k_1,k_2-1)\vee X(k_1,k_2)$. And this cofiber is isomorphic to a wedge of $(k_2-k_1)$-dimensional spheres, by the proof of Claim 2. So up to homotopy $X(k_1,k_2)$ is a retract of a wedge of such spheres. I suppose you could also figure out the rank of $\tilde H_{k_2-k_1}X(k_1,k_2)$ by this method, by induction on $k_2$.
