Given that $g: \mathbb{N}\rightarrow \mathbb{N}$ such that for all $a,b \in \mathbb{N}$, if $a < b \Rightarrow g(a) < g(b)$.
I want to show that for all $x \in \mathbb{N}$, we have $x \leq g(x)$.
I use proof by induction.
Base case: Let $x=1$, then $1 \leq g(1)$.
Assume $x=k$ by Mathematical Induction: $k \leq g(k)$ for all $k \in \mathbb{N}$.
We want to show that $x=k+1$, i.e., $k+1 \leq g(k+1)$.
I start from $$k \leq g(k)$$ Adding $1$ both sides gives, $$k+1 \leq g(k) + 1$$ Since $g(k) < g(k+1)$, then $$k+1 < g(k+1) + 1.$$ So $$k<g(k+1).$$ Since $k<k+1$, $k+1\leq g(k+1)$ for all $k \in \mathbb{N}$.
Does this proof make sense?
Note there is also a possibility that the statement can be wrong, but I think it's correct since $g$ is a monotonically increasing function. Had a hard time finding a counterexample, but that doesn't mean the statement is true.
Yes, your proof is correct.
You have proved $ k<g(k+1)$ and since g(k+1) is an integer which is bigger than $k$ it must be greater or equal to $k+1$ that is $k+1\le g(k+1)$