Is this proof of strong induction valid?

Question

Let $n, m$ and $m_0$ be natural numbers such that $n>m_0$. Let $Q(n)$ be the property that $P(m)$ is true if $n>m≥m_0$. If $Q(n) \implies P(n)$, then we can conclude that $P(m)$ is true of all $m≥m_0$.

Now, the proof:

We induct on $n$. Assume that, if $P(m)$ is true for each $n>m≥m_0$, then $P(n)$ is also true. Since $n+1>n$, we know that $n≥m_0$. Thus, we have that $P(n)$ holds for all $n+1>n≥m_0$. Then, by inductive hypothesis, $P(n+1)$ is also true. This closes the induction, and we can conclude that $P(n)$ is true for all $n≥m_0$.

Answer 1

Supposed I have the following proof:

Claim: $n^2-2>0\forall n$ natural

Proof: assume that the above claim is true I get $(n+1)^2-2=n^2+2n-1>n^2-2>0$

Can you see what is wrong here?

I proved that if my claim is true for some $n$ it is true from all above it, but I never proved it to $n=1$(which is not true for).

In your case, you have to prove the assumption that it is true for $n>m\ge m_0$, or at least for $m=m_0$.