In order to apply induction, we generalize the result to be proved so that it reads as follows:

Proposition. If the $n$-element subsets of a set $S$ with $(n+1)m-1$ elements are partitioned into two classes, then there are at least $m$ pairwise disjoint sets in the same class.

Proof. Fix $n$ and proceed by induction on $m$. The case of $m=1$ is trivial. Assume $m>1$ and that the proposition is true for $m-1$. Let $\mathcal{P}$ be the partition of the $n$-element subsets into two classes. If all the $n$-element subsets belong to the same class, the result is obvious. Otherwise select two $n$-element subsets $A$ and $B$ from different classes so that their intersection has maximal size. It is easy to see that $|A \cap B| = n-1$. (If $|A \cap B| = k < n-1$, then build $C$ from $B$ by replacing some element not in $A \cap B$ with an element of $A$ not already in $B$. Then $|A \cap C| = k+1$ and $|B \cap C| = n-1$ and either $A$ and $C$ or $B$ and $C$ are in different classes.) Removing $A \cup B$ from $S$, there are $(n+1)(m-1)-1$ elements left. On this set the partition induced by $\mathcal{P}$ has, by the inductive hypothesis, $m-1$ pairwise disjoint sets in the same class. Adding either $A$ or $B$ as appropriate gives $m$ pairwise disjoint sets in the same class. $\square$