You seem to have the right idea. Let me address a few issues of execution.
Assume $\mathbb{N}\sim B$ by a function $f$.
1) $A \subseteq B$ means that $\forall a \in A, a \in B$
If we use the same function $f$ that defines $\mathbb{N}\sim B$ and apply it to $N_A = \{ n : f(n) \in A \}$, then $\exists a_1, a_2 \in A : a_1 \ne a_2 \Rightarrow f(a_1) = f(a_2)$ this would lead to $\mathbb{N} \nsim B$ because $a_1,a_2 \in B$ by (1). So any function $f$ that is $1:1$ from $\mathbb{N}$ to $B$ must be $1:1$ from $N_A$ to $A$.
One critique about this part: the domain of $f$ is $\Bbb N,$ but if $A\ne B,$ then $N_A$ will not be equal to $\Bbb N$! Hence, $f$ is not a function on $N_A$ in such a case. Rather, the restriction of $f$ to $N_A$ is the function you want. In particular, let $g:N_A\to B$ be defined by $g(n):=f(n)$ for all $n\in N_A.$ Then you can readily adapt these arguments to show that $g$ is one-to-one, and in fact, $g:N_A\to A.$
Onto is proven in the same way. $\nexists n \in N_A : f(n) = a, \forall a \in A \Rightarrow \mathbb{N}\nsim B$ because of (1).
I am having a difficult time parsing this logic language in a way that makes sense. Simple verbal argument is a better approach, here.
If $a\in A,$ then $a\in B$ by (1), so by definition of $N_A$ there is some $n\in N_A$ such that $f(n)=a,$ and by definition of $g$ we have $g(n):=f(n)=a.$ Thus, $g:N_A\to A$ is onto, and so $N_A\sim A.$
$N_A \subseteq \mathbb{N}\Rightarrow$ $A$ is countable.
Absolutely not, since it seems like your definition of "countable" is "in bijection with $\Bbb N.$" After all, $N_A\subseteq\Bbb N$ by definition. It could be empty or finite, remember?
$N_A = \emptyset \Rightarrow A = \emptyset$.
Very true!
$\exists s \in \mathbb{N} : s = sup (N_A)\Rightarrow A$ is finite because $\forall n \in \mathbb{N}, n > t, f(n) \notin A$
I think you meant $s,$ rather than $t,$ but with that fix, this is fine.
Now, it remains to prove the last case. Namely, suppose that $N_A$ has no upper bound--that is, that for all $m\in\Bbb N,$ there is some $n\in N_A$ such that $n>m.$ You must show that $\Bbb N\sim N_A.$ To do so, consider the recursively-defined function given by $h(b):=\min\left(N_A\right)$ (where $b$ is the least element of $\Bbb N$) and $h(n+1)=\min\left(N_A\cap\{m\in\Bbb N:m\le n\}\right).$ You should be able to show that this $h$ is well-defined, and that $h:\Bbb N\to N_A$ is one-to-one and onto, completing the proof. (As a side note, this is where induction comes into play, though possibly in a disguised form.)
