Is this horse proof by induction okay? Let $P(n)$ be the statement "all horses in a set of n horses are of the same colour."
Basis Step: Clearly, $P(1)$ is true.
Inductive Hypothesis: Suppose that $P(k)$ is true for some arbitrary integer $k\geq 1$; that is, all horses are of the same colour.
Inductive Step: We now prove that $P(k+1)$ is true.
Consider any $k+1$ horses. Number these horses as $1,2,3,...,k+1$. By the inductive hypothesis, horses $1,2,...,k$ have the same colour. Also, by the inductive hypothesis, horses $2,3,...,k+1$ have the same colour. Because the set of the first $k$ horses and the last $k$ horses overlap, all $k+1$ horses must be of the same colour and we have shown $P(k+1)$ is true.
Therefore, all horses in a set of $n$ horses are of the same colour, for all integers $n\geq1$.
 A: Have you read this? It's from Wikipedia.

Explanation:
  The argument above makes the implicit assumption that the two subsets of horses to which the induction assumption is applied have a common element. This is not true when n = 1, that is, when the original set (prior to either removal) only contains n+1 = 2 horses.
  Let the two horses be horse A and horse B. When horse A is removed, it is true that the remaining horses in the set are the same color (only horse B remains). If horse B is removed instead, this leaves a different set containing only horse A, which may or may not be the same color as horse B.
  The problem in the argument is the assumption that because each of these two sets contains only one color of horses, the original set also contained only one color of horses. Because there are no common elements (horses) in the two sets, it is unknown whether the two horses share the same color. The proof forms a falsidical paradox; it seems to show something manifestly false by valid reasoning, but in fact the reasoning is flawed

