Proving infinite subsets I have to prove: 
An infinite subset of a denumerable set is denumerable. 
I understand this has been asked before and I did take the time to read what was said there, but I do not understand still. 
I have to prove this using other theorems about denumerable or countable sets, nothing too complex for this proof. 
 A: Suppose that the subset is not a countable infinite set, so if you prove that your subset $A$ must be more than countable in order to be infinite you have done, because then it can't be a subset of you denumerble set. Suppose that there exists an infinite set $B$ whose cardinality is less than the cardinality of the natural numbers that is infinite; so there exists an injective function $f: B \rightarrow \mathbb N$: its image is infinite by hypothesis, and it is a subset of the natural numbers so it has a minimum $x_0$: associate this minimum with $0$ and then consider the set $f(B)\setminus x_0$ and iterate considerating the new minimum $x_1$; your process never ends because $f(B)$ is infinite. So we have shown that there is a bijection between $B$ and $\mathbb N$, that is what we wanted to prove. 
A: One of the many possible proofs:
If a set $A$ is denumerable, then there is a bijection $f:A\rightarrow \mathbb N$
So if $B\subseteq A$,the restriction of $f$ to $B$, which is $f:B\rightarrow \mathbb  N$ is $1:1$
On the other side, if $B$ is infinite, we can construct an $1:1$ function $g:\mathbb N\rightarrow B$ by defining $f(0)=x_0\in B$, then $f(1)=x_1\in B-\{x_0\}$, and so on by induction.
So by Schröder–Bernstein there is a bijection between $B$ and $\mathbb N$, which makes $B$ denumerable.
