How many ways are there to break the people into two teams of two? Consider a group of four people.
(a) How many ways are there to choose a two-person committee?
(b) How many ways are there to break the people into two teams of two?
I could solve (a). to form a two person committee from a group of four people, the first position could be taken by 4 people and second position by 3 people. so 4 times 3 is 12 and to account for double counting we divide by 2. So we can choose a two person committee in 6 ways.
I am not able to make sense of (b).
 A: Suppose the teams are labeled, say team $A$ and team $B$.  Then there are $\binom{4}{2}$ ways to select the two members of team $A$ from among the four available people.  The other two people must comprise team $B$.  Thus, if the teams are labeled, there are $\binom{4}{2}$ ways to form the teams.
However, you will notice that the problem does not say that the teams are labeled.
If the teams are not labeled, a team is determined by its members.  Observe that the two teams of two people are completely determined by which of the other three members is paired with the oldest member of the group.  Thus, there are three ways to form two unlabeled teams.
A: There are $4$ choices for the first team member. Then, there are $3$ choices for the second person. This means there are $12$ possible groups. To get rid of double counting, we divide by $2$ to get $6$.
A: b) There are $C(4,2)$ ways to get two persons from four people. This is the first team. The second team is consisted of the other two people. The number of ways, thus, is $C(4,2)=\frac{4!}{(4-2)!2!}=6$.
If the teams are indistinguishable, then divide this result by 2: $\frac{6}{2}=3$.
A: Suppose that you have to choose first team. So you choose two members from all 4. So you get $\binom{4}{2}$. For the next team, you need to choose another 2 members. But, now you will have to choose them from 4 members (initialy) - 2 members (first team) = 2 members, so you will get $\binom{2}{2}$. So the number of ways you can form the teams are $\binom{4}{2} \binom{2}{2} = \frac{4!}{2!(4-2)!}*\frac{2!}{2!(2-2)!}=3*2=6$.
A: Person 1 has to belong on some team, and whatever team they are on there are only two people left for the other team, so once their partner is chosen the teams are set.  Person 1 can be with any of the other three people, and the remaining two will be on the other team.  So there are 3 ways to make two teams of two out of four people.
In your example when you choose a two person committee, the other two people are also a valid match-up.  Due to this double-counting, you can divide 6 by 2 to get 3.
A: The formulation of the second question (to me) clearly asks for the number of partitions of the set of the $4$ people into parts of size$~2$ each. A partition of a set$~S$ is a set of non-empty subsets that are mutually disjoint and whose union is$~S$. So one partition of $\{a,b,c,d\}$ into parts of size$~2$ each is $\{\{a,c\},\{b,d\}\}$. Since the order in which the elements of a set are listed is of no importance, this is the same partitions as $\{\{c,a\},\{b,d\}\}$, and also as $\{\{b,d\},\{a,c\}\}$.
From part (a) you know that there are $6$ subsets each containing $s$ of the $4$ people. Each such subset determines a partition by grouping the remaining two people into another subset and taking both groups as the two teams. But every partition into two groups of $2$ is obtained twice in this way, once for each of its two teams being chosen first. TO get the correct number of partitions you must divide by this factor $2$, to get the correct result $6/2=3$.
More generally if one wants to count the number of partitions of a set $n$ into parts of prescribed sizes $k_1,k_2,\ldots,k_l$, with $k_1+k_2+\cdots+k_l=n$ so that this is possible at all, then one can first count the number of ways to color the $n$ elements with $l$ (team-)colors, so that for each $i$ exactlt $k_i$ of them get color $i$, which is known to be the multinomial coefficient
$$ \binom n{k_1,k_2,\ldots,k_l} = \frac{n!}{k_1!\,k_2!\ldots k_l!}$$
But then, to get the number of partitions of this type, you must forget about the actual colour assignments, and just retain the subsets defined by them. This means that whenever multiple numbers $k_i$ are equal, we can permute the corresponding colours in any way among each other without changing the partition defined by the colouring, and if this happens for different group sizes the corresponding permutations of colours can be applied independently. Therefore for every value that occurs multiple times as a group size$~k_i$, say $m$ times in all, one must divide the number by another factor$~m!$.
The exercise in the question is the simplest case where this adjustment in the computation of the number of set partitions must be applied (which is probably the purpose of the exercise); on has $n=4$ and $(k_1,k_2)=(2,2)$, and the formula becomes
$$ \binom4{2,2}/2!=\frac{4!}{2!\,2!}/2!=6/2=3.
$$
