Does you see why the overcounting happens? I'll attempt to explain.
Suppose your people are A,B,C,D,E,F,G,H,I,J,K,L. You are picking teams. Let me use your algorithm:
1) You pick two people first, say A and B. So the first group is (A,B).
2) You pick five people next, say C,D,E,F, and G. So the second group is (C,D,E,F,G).
3) the leftovers form a group, so that is (H,I,J,K,L).
So the three groups are (A,B), (C,D,E,F,G) and (H,I,J,K,L).
But now, suppose you did the following:
1) You pick two people first, say A and B. So the first group is (A,B).
2) You pick five people next, say H,I,J,K,L. So the second group is (H,I,J,K,L).
3) the leftovers form a group, so that is (C,D,E,F,G).
So the three groups are (A,B), (C,D,E,F,G) and (H,I,J,K,L) : AGAIN!
Hence the overcounting happens, because you confuse the two groups of five while picking them in an order. And that's why you divide by $2$, because you know you are only to confuse a group of five with another group of five, so the two get mixed up, and you divide to two to negate the order in which the groups are picked.
In general, it is not wrong to overcount. However, it is important to see where order comes into play, and where simply grouping comes into play. When you are picking objects and arranging them, then you are desiring order. However, when you are picking groups, then neither does the inter group arrangement matter nor does the intra group arrangement matter. What I mean to say is , the arrangements (A,B) and (B,A) is an intra arrangement of the group, which we avoid by taking combinations instead of permutations. However, $(A,B),(C,D,E,F,G),(H,I,J,K,L)$ is an inter arrangement of groups, and one that differs from $(A,B),(H,I,J,K,L),(C,D,E,F,G)$ according to your logic, because this involves the order of picking, which in truth is irrelevant. To avoid this discrepancy, we divide by $2$, because this gets rid of the inter arrangement mess. Thus over counting takes place.
Let me phrase you questions for practice :
Suppose you have 14 people. How many ways are there to divide them into seven groups of two each?
Suppose you have 65 people. How many ways are there to divide them into seven groups of seven and four groups of four?
Either get back with the answer, because if you get it right, you are on the right track. Always devise an algorithm, and then think about avoiding inter and intra grouping.
Best Answer
You are correct in saying that the first person can be chosen in $6$ ways, the second in $5$ & the third in $4$ ... BUT
This group of $3$ people could have been chosen $3!$ ways ... SO
there will be $6 \times 5 \times 4 /3!$ ways.
Another way to think about it is ... how many ways could you choose $3$ people from $6$ ... $\binom{6}{3} = \color{red}{20}$.