If set A is equivalent to a subset of N then is A countable? There’s not really much context to this. Is this connected to the fact that if A is equivalent to the set of natural number then A is countable(more specifically countably infinite)?
 A: Yes, and it is connected to the countability of the naturals. We can more precisely claim a set $S$ is countable if there exists an injection $S \to \Bbb N$ (or to any other countable set, e.g. $\Bbb Z, \Bbb Q$). For clarity, we say $S$ is countable if $|S| \le \aleph_0$.
More generally, if there exists an injection $A \to B$, then we have that $|A| \le |B|$. (Similarly a surjection would have $|A| \ge |B|$, assuming the Axiom of Choice, and a bijection would have $|A|=|B|$.)
A: Following the lead of @EeveeTrainer, we can use following definition:
Countable set: Let $A$ be a set. $A$ is countable iff there exists an injection $c:A\rightarrow \mathbb{N}$.
Let $A$ and $B$ be sets such that $A\cong B$. Recall that this means there exists an injection $i:A \hookrightarrow B$ (which is also a bijection). Now, if there exists an injection $c:B \hookrightarrow \mathbb{N}$ i.e. if $B$ is countable, we can compose maps $c$ and $i$ to get an injection $c \circ i:A \hookrightarrow \mathbb{N}$, thereby proving that $A$ is also countable. (Notice I used the fact that composition of two injections results in an injection). 
As an aside, note that this does not require B to be a subset of natural numbers in the membership-based set theory sense.
