If $f : A \to B$ and $B$ is countable, given that $f$ is surjective, $A$ is countable.

I'm attempting to prove this statement, and I'm not sure if my deductions make sense, so I'd very much appreciate it if anyone could critique, comment or (most probably) improve my train of thought.

Since we're given that $B$ is countable, then by definition there must exist some $g$ s.t. $g : \mathbb{N} \to B$, with $g$ surjective. Then, it must be the case that $A$ is similarly countable because since $B$ is countable, and $f : A \to B$, for this to hold it's necessary that $A\subseteq \mathbb{N}$. Because $\mathbb{N}$ is countable, then $A$ is countable.

Again, I'm definitely not sure that is correct, I'm still getting my head around how exactly I'm supposed to prove statements like these, so any criticism would be amazing.

This statement is incorrect. A counterexample is $A = \mathbb{R},\, B = \mathbb{Z}$, and $f(x) = [x] = \max \{ z \in \mathbb{Z} \, : \, z \le x\}$.

As to critique - your proof attempt "Then, it must be the case that A is similarly countable because since B is countable ..." consists mainly of restating what is to be proved.