How many ways are there to split $10$ people into two groups of $5$?

In the first one, there is a difference between the two resulting teams. One is playing, the other is sitting on the sidelines. So swapping the two teams around makes for a different choice of starting five.

In the second case, there is no difference. The only thing that matters is who is on team with who, and not which team is team A and what team is team B. So swapping the two teams around still gives the same choice of two teams.

You can also go the other way: Choose a starting five by first dividing into two teams (which can be done in some number of ways), then choosing which of the two teams gets to be athe starting five. You have two options for starting five for each division into two teams. Thus you have to multiply by 2 to get from one to the other.

Observe that the choice of choosing five people and the other five are counted separately in the first case, but yield the same separation into two groups in the second. That's why there is division by $2$.

Note that when we choose $\binom{10}{5}$ we are counting twice the same teams, as for example

$$ \{ A,B,C,D,E \} , \{ G,H,I,L,M \} \equiv \{ G,H,I,L,M \},\{A,B,C,D,E\}$$

therefore the number of ways to obtain 2 teams is $\frac12 \binom{10}{5}$.