I am so sorry for posting about it more than one time but this is my 4th revision for this proof and I want some feedback..
Note : $[n]$ is $\{1,2,3,4,\ldots,n\}$, a finite set
Prop: Every subset of a finite set is finite
Proof: Let $A,B$ sets and suppose $B$ is a finite set.
Also let $A \subseteq B$
If $A = \varnothing$, then it is finite
If $B = \varnothing$, then also $A = \varnothing$, which is finite.
So we suppose $A \neq \varnothing$ and $B \neq \varnothing$. Then there exstis a bijection $f: B \to [n]$
Since $A \subseteq B$, the inclusion mapping $I : A \to B$ defined on $f(a) = a$ for $a \in A$ exists.
Note $I(a_1) = a_1$ and $I(a_2)=a_2$ and if we suppose $I(a_1)=I(a_2)$, then $a_1 = a_2$, and therefore $I$ is an injection.
Then we can have $f \circ I: A \to [n]$, which is injective.
We need to prove that there is a bijection between the set $A$ and subset of $[n]$
So, let $p(n)$ be the statement that "If $A \subseteq [n]$, $A$ has a bijection with $[m]$ for some $m \in Z_{\ge0}$ such that $m \le n$"
Base case $(n=1)$ : If $A \subseteq [1]$, since we suppose $A \neq \varnothing$, $A$ has a bijection with $[1]$
Induction part : Now let $k \in N$ and suppose $p(k)$ holds. Now we want to prove that $p(k+1)$, that is, "If $A \subseteq [k+1]$, $A$ has a bijection with $[m]$ for some $m \in Z_{\ge0}$ such that $m \le k+1$"
Then if $A $ has $k+1 $ elements, it is done. It has a bijection with $[k+1]$.
If $A \subseteq [k]$, by the induction hypothesis, $A$ has a bijection with $[m]$ for $m \le k$.
Hence every subset of finite set is finite.
Best Answer
Would this be easier? An infinite set $S$ trivially has infinitely many subsets (take e.g. the singletons $\left\{ s \right\}$, $s \in S$). By a simple combinatorial argument a finite set $F$ has $2^{\#F}< \infty$ subsets.
Assume that $A \subset F$ is infinite, with $F$ a finite set. Then there are infinitely many $B \subset A \subset F$, which is a contradiction.