Finding the mistake in induction proof This argument seems obviously wrong, but I can't find the mistake. Any ideas?

There are two people, each given a positive integer, such that the two integers differ by $1$. The rule of the game
  requires that if anyone knows that if his number is the smallest among the two, then he should put up his hand. We
  are going to prove that the one with the smaller number will always put up his hand.
Suppose the two people are
  given positive integers $n$ and $n + 1$. We shall prove that the one given the integer $n$ will put up his hand. Note that
  this is true for $n = 1$ since $1$ is the smallest positive integer, so the one given the number $1$ must know that his number
  is the smaller of the two.
Suppose the statement holds for some $n$ and that the two people are given $n + 1$ and $n + 2$.
The one with $n + 1$ will think
  “the other number is either $n$ or $n + 2$.” Suppose that it is $n$. Then, by induction hypothesis, the other person should
  put up his hand. But he doesn’t, so the other number must be $n + 2$.
Consequently, the one with $n + 1$ will put up his hand.

 A: The induction is not valid as it is not well defined. What I mean, is that the "waiting time" need not be the same all the time. 
For example, under the assumption that a person takes 1 second to raise his hand, if player B had the number 2, he would wait 1 second, and if his partner does not raise his hand, then he will raise his. The same reasoning applies if he had the number $n$. If a person does not raise his hand after $n-1$ seconds, then it would be clear that he would not have the number $n-1$.
However, this is not the case in your problem, generating a kind of paradox similar to this one. Again, the problem here is that in your proof it should say "Then, by induction hypothesis, the other person should put up his hand  after $X$ time. But he doesn’t." As you do not define $X$, you get that paradoxical result.
A: The base case is proven for the numbers $1$ and $2$ (denoted $n$ and $n+1$).
The induction works with $n+1$ and $n+2$ instead, hence the base case should correspond to $\color{red}{n=0}$. Then the statement 'The one with $n+1$ will think "the other number is either $\color{red}n$ or $n+2$"' does not hold.
This is a variant of https://en.wikipedia.org/wiki/All_horses_are_the_same_color.
