Call the set containing the sets of intervals $S$ and build a graph $G_S$ from $S$ as follows: Each set of intervals $I \in S$ becomes a vertex, and there is an edge between interval set $I$ and interval set $J$ if and only if some interval in $I$ overlaps with some interval in $J$. So for your example, the graph would have three vertices:

$$ A = \{(1,2),(3,4)\}, B = \{(1,3)\}, C = \{(13,14)\}$$

And there is an edge from $A$ to $B$ since $(1,2)$ overlaps $(1,3)$.

Now, the number of connected components of $G_S$ gives you the "number of non-pairwise overlapping interval sets" and picking a vertex from each connected component furnishes a solution to your problem.

So back to your example: the connected components are $AB$ and $C$, so you can pick either $A$ and $C$ or $B$ and $C$, as you have said.

Regarding efficiency: building this graph is $O(m^2n)$ where $m$ is the cardinality of $S$ and $n$ is the maximal cardinality of any interval set $I \in S$.

Call the set containing the sets of intervals $S$ and build a graph $G_S$ from $S$ as follows: Each set of intervals $I \in S$ becomes a vertex, and there is an edge between interval set $I$ and interval set $J$ if and only if some interval in $I$ overlaps with some interval in $J$. So for your example, the graph would have three vertices:

$$ A = \{(1,2),(3,4)\}, B = \{(1,3)\}, C = \{(13,14)\}$$

And there is an edge from $A$ to $B$ since $(1,2)$ overlaps $(1,3)$.

Now, the number of connected components of $G_S$ gives you the "number of non-pairwise overlapping interval sets" and picking a vertex from each connected component furnishes a solution to your problem.

So back to your example: the connected components are $AB$ and $C$, so you can pick either $A$ and $C$ or $B$ and $C$, as you have said.

Regarding efficiency: the worst-case complexity of building this graph is $O(m^2n^2)$ where $m$ is the cardinality of $S$ and $n$ is the maximal cardinality of any interval set $I \in S$.