Subset of Finite Set is Finite

Theorem
Let's recall that a set $X$ is finite when it's either empty, or there is a natural number $n\in\left\{ 1,2,3,\ldots\right\}$ and a bijection $f:\left\{ k\in\mathbb{N}:k\leq n\right\} \to X $.

This theorem says that if Y is a subset of X, then Y is also finite.

Lemma
Let $f:A \to B$ be bijective. Then given $a\in A$ and $b\in B$, we can find another bijection $g:A \to B$ such that $g\left(a\right)=b $.

Proof of Lemma
Let $b'=g\left(a\right)$. Since $f$ is surjective, there is $a'\in A$ such that $f\left(a'\right)=b$. Let's define $g:A\to B$ by putting $g\left(a\right)=b$, $g\left(a'\right)=b'$, and $g\left(x\right)=f\left(x\right)$ for $x\in A\setminus\left\{ a,a'\right\}$.

Well, $g$ is clearly surjective, for if we choose $y$ either in $B\setminus\left\{ b,b'\right\}$ or in $\left\{ b,b'\right\}$, we can find $x$ such that $f\left(x\right)=y$. Using a similar proof structure, one can show that $g$ is also injective, hence it's bijective.

Proof of the Theorem
We will begin proving the following particular case: if $X$ is finite and $a\in X$, then $X\setminus\left\{ a\right\}$ is also finite.

Well, there is a bijection $f:\left\{ k\in\mathbb{N}:k\leq n\right\} \to X$, which we can suppose that satisfies $f\left(n\right)=a$. If $n=1$, $X\setminus\left\{ a\right\}=\emptyset$ is finite.