Theorem: Principle of Mathematical Induction
For each natural number $n$, let $P(n)$ be a statement. If
$P(1)$ is true and
$P(k) \Rightarrow P(k+1)$ for every $k \geq 2$
Then $P(n)$ is true for all $n$.
Shouldn't that be $k\geq1$?
Otherwise, I cannot logically conclude $P(2)$'s truth from anything, and thus the chain of modus ponens cannot even be started.
In your formulation of the inductive step 2), it must be $k\ge 1$. You can use $k\ge 2$ if you write $$ 2) \quad P(k-1) \Rightarrow P(k) $$
As suggested by the other posters, the statement is wrong, but I want to address Git Gud's idea.
Claim: No natural number is equal to $2$. $(P(n) \iff n \neq 2 )$
We will prove this theorem using the principle of broken induction. First, we see that $P(1)$ is true, since $1 \neq 2$.
Next, we show that $P(k) \Rightarrow P(k+1)$ for all $k \geq 2$.
Case #1; suppose that $k = 2$. Then $P(k)$ is false, and thus $P(k) \Rightarrow P(k+1)$ is vacuously true.
Case #2; suppose that $k > 2$. Then $P(k)$ is true, and $k + 1 > 2$. Thus, $k+1 \neq 2$, so $P(k+1)$ is true. Thus, $P(k) \Rightarrow P(k+1)$ is true.
Since $P(k) \Rightarrow P(k+1)$ holds for all $k \geq 2$, and $P(1)$ has been shown to be true, the proof is complete.
Thus, no natural number equals $2$.