Why the following induction proof is wrong?
Claim: Natural numbers $0,1,2,3,\dots$ are all even.
Proof by strong induction:
Base case: $n=0$ is an even number, hence the statement is true for $n=0$.
Inductive step: Assume that the statement is true for $n=0,1,2,\dots,k$, and consider $n=k+1$. By assumption, both 1 and $k$ are even numbers, and hence so is their sum $k+1$. It thus follows that the statement holds for all $n=0,1,2,3,\dots$
The following statement is not true for $k = 0$
"By assumption, both $1$ and $k$ are even numbers, and hence so is their sum $k + 1$."
For $k = 0$, you only know that $0$ is even not $1$. So you can't make the assertion above for all $k$.