Prove that Set B is countable - Is this proof correct? It seems that I have some issues with the rigor of this proof and I don't know what I'm doing wrong. Could someone tell me if this proof is correct and rigorous enough?
Here's the question

Prove that if B$\subseteq$A and A is countable, then B is countable.

Here's how I proved it:

Suppose B$\subseteq$A and A is countable. Since A is countable, it is
finite or denumerable. Considering cases:
Case 1: A is finite. Then A
$\sim I_n$ where n=|A|. Since B$\subseteq$A, it follows B $\sim I_n$.
So B is finite. Therefore B is countable.
Case 2: A is  denumerable. $Z^+ \sim$ A. Since B$\subseteq$A, it
follows $Z^+ \sim$ B. So B is denumerable. Therefore B is countable.
Thus, B is countable as desired.

**$Z^+$ is the positive integers.
Any help would be much appreciated. Thanks.
***Update:some clarifications,
$\sim$ means equinumerous
A being countable means A is either finite or denumerable.
A set A is called denumerable If $Z^+$ $\sim$A.
 A: When you say "Since $B\subseteq A$ it follows $B \sim I_n$" one could consider that merely stating that which was to be proved by inserting "it follows" -- not really a proof.
I would start with the fact that since $A$ is countable there exists a "counting function $C(A)$" (an injective map $C : A \mapsto \Bbb{N} : \forall a, b \in A, C(a) = C(b) \implies a=b $.  Now consider the restriction of that counting function to the the subset $B$, and it is easy to show that this is an injection as well, thus a valid counting function for $B$.
A: I'm not terribly familiar with the $\sim$ notation for sets; presumably it means they're in bijection? 
At any rate, I think the difficult part of the argument involves infinite subsets of $A$. Consequently, I think you could use more of an argument in this case, rather than "it follows that $Z^+ \sim B$." 
For example, if it's because $B \subseteq A$ gives rise to an injective inclusion function, and the composition of injective functions is injective, then say so!
