Unboundedness of infinite subsets of natural numbers

$\newcommand{\N}{\mathbb{N}}$ I am trying to show that every infinite subset of $\N$ is unbounded, that is, $$ \forall A \subseteq \N: (|A| = |\N| \Rightarrow \forall m \in \N:\exists n \in A:m < n) $$

It seems obviously true because, informally, the negation of the consequent results in the finiteness of $A$, which is a contradiction.

However, I failed to formally show this fact using the definition of equipollence: $|A| = |\N|$ iff there is a bijection from $A$ to $\N$. Could you give comments on this problem?

Edit: It would be sufficient to show that there is an injection $A \to \N$ but no surjection from the negation of consequent: $\exists m \in \N: \forall n \in A: n \le m $.

Solution 1:

Define $f: \Bbb N \to A$ by induction. Set $f(0)$ to be the least element of $A$, and then set $f(n+1)$ to be the least element of $A$ above $f(n)$. This enumerates $A$ in increasing order and $f(n) \geq n$.

If $A$ were bounded, this process would give you a surjection from a finite number to $A$, which isn’t possible for infinite $A$.