Are proofs by "maximality" equivalent to proofs by induction? I apologize for the lack of proper terminology; I have zero experience in this field.

What I mean by "proof by maximality": One way to show that a set $A$ has a certain property $p$ is to assume there is a largest proper subset $X \subsetneq A$ which verifies $p$ and then show that there is another subset of $A$ larger than $X$, $Y\supsetneq X$, which also verifies $p$, a contradiction. Then one concludes that $A$ verifies $p$.

Now, these kinds of proofs seem strikingly similar to proofs by induction, where one assumes that a proposition holds for $n$ (i.e. $n$ is the largest number that verifies it) and then proves that it must hold for $n+1$ as well and therefore, by the principle of induction, it is true for all the elements of $\mathbb{N}.$ The "base case" is similar as well, since in proofs by "maximality" one first has to show that there exists a subset verifying the property $p$.

I was wondering if there is a link between these two proof techniques and, if at all, how one can be turned into the other. I am having trouble trying to formalize this. Where can I learn more about it?
Well, not familiar with your argument. But the induction axiom goes in the direction of maximality.

Let ${\Bbb N}_0$ be the set of natural numbers. Let $N$ be a subset of ${\Bbb N}_0$ such that $0\in N$ and with each $n\in N$ we have $n+1\in N$. Then $N = {\Bbb N}_0$.
Well, not familiar with your argument. But the induction axiom goes in the direction of maximality.

Let ${\Bbb N}_0$ be the set of natural numbers. Let $N$ be a subset of ${\Bbb N}_0$ such that $0\in N$ and with each $n\in N$ we have $n+1\in N$. Then $N = {\Bbb N}_0$.