Let $A$ be a finite set. If $X \subseteq A$, then $X$ is finite.
Proof:
Since $A$ is finite, there is some $n\in N$ such that $A$ ~ $[1,n]$ and there is some bijective map $f:A \to [1,n]$. Let's define a map $g : X \to f(X)$ such that $g(x)=f(x)$ for all $x \in X$. Since $f$ is injective, so is $g$, and since the codomain of $g$ is $f(X)$, from the definition of image, $g$ is surjective and hence $g$ is bijective. Since $f(X) \subseteq [1,n]$, so there is some $k \in N$ such that $f(X)$ ~ $[1,k]$, and hence $X$ is finite.
Note that the symbol ~ means that the the sets have the same cardinality
So is there any flaw in the proof or some point that is not exactly clear or maybe unnecessarily explained etc. I'm hoping you guys can check the proof and give me some feedback about it in any aspect of the proof.