Proving that $A$ is countably infinite from another statement

Question

I am trying to learn Real Analysis from lectures available online: Mathematics-Real Analysis(nptel)

The professor tries to prove that (3) implies (1) in the following (in his 4th Lecture of 1st module):

1. $A$ is countably infinite,
2. $\exists$ a subset $B$ of $\mathbb{N}$ and a map $f:B \to A$ that is onto,
3. $\exists$ a subset $C$ of $\mathbb{N}$ and a map $g:A \to C$ that is one-one.

Proof: (At 22:00 of the above video)

Consider the map $A\to g(A)$

Now, this is onto and hence, $A\approx g(A)$

also, it is clear that $g(A)\subseteq \mathbb{N}$ and we know that every subset of $\mathbb{N}$ is countable hence, g(A) is countable and particularly countably infinite

therefore, $A$ is countably infinite

My Doubt: why $g(A)$ is countably infinite ?

Answer 1

3) implies 1) is false. $C$ and $A$ could both have one element each in which case there is a one-to one map from $A$ to $C$.