Inductive proof that every person supports the same team I can't find what's wrong with this.
Proof:
Every person supports the same team.
We can observe that in a set with only one person, all people within it support the same team. Now suppose that the statement is true for every set with cardinality $≤ n$. Then if there are $n + 1$ persons in a set, we take one of them and, by the hypothesis, all $n$ persons support the same team. Now put this person back to the initial set and remove a different one. Again, all the $n$ persons left support the same team. Therefore all $n + 1$ persons support the same team, and for every $k$ in $\mathbb N$, those $k$ people support the same team.
 A: 
Now put the person back and remove another.  Again those n people support the same team. 

Yes, but the $n$ people might support a different team than the previous $n$ people.  
You are assuming:
$ set_1 = \{$ $n$ people supporting team $A$ $\}$
$ set_2 = \{$ $n$ people supporting team $B$ $\}$
$ set_{phantom} = set_1 \cap set_2 = \{$ $n-1$ people who support both team $A$ and team $B$ $\}$
(NEVER mentioned but assumed to exist)
We assume $set_{phantom}$ isn't empty and as people can't support different teams, we conclude $A$ and $B$ are the same team.
Then we have:
$set_{magic} = set_1 \cup set_2 = \{ n + 1$ people supporting either team $A$ or team $B$ but they are the same team $\}$
But notice! Our base case was $n=1$.  For $n=1$ then $set_{pantom}$ with $n-1 = 0$ members IS empty. So team $A$ and team $B$ may be different. And $set_{magic}$ could be $n+1 =2$ people supporting different teams.
So to do this proof right we need to prove it for a case where $set_{phantom}$ will have at least $1$ member. Thus we need to show this for a base case of $n >= 2$.
So we need to show all sets of two people support the same team.  
It is impossible for us to show this.
A: Hint: The base case needs to be solved for $n = 1$ AND $n = 2$ in this problem. Do you see why?
