Show that maximum disjoint set problem is NP Complete
Given MIS is NP-Complete, we can reduce MIS to MDS. Let $G = (V,E)$ the graph we have for MIS problem. For every vertex $v_i \in V$, create a set $S_i$ with edges of $v_i$, however, be careful that write each edge element as $v_iv_j$ even if it is $v_jv_i$. Reduction is completed after creating the sets $S_i$. There is an independent set in $G$ of size $m$ if, and only if, there are $m$ disjoint sets where sets are taken from independent set's vertices.

This reduction can be performed in polynomial time since we traverse each vertex $v$ and its edges. It can be performed in $O(V+E)$. Additionally, given a collection of sets, we can verify if they are disjoint and there are at least $k$ of them in polynomial time.

Consequently, MDS is NP-Complete