Prove that a set is infinite if and only if it is equipotent to a proper subset. Prove that a set $X$ is infinite if and only if it is equipotent to a proper subset $H$.
Proof.
Suppose $X$ is equipotent to a proper subset $H$. Then there exists a bijection $f: H \to X$. Toward a contradiction, suppose $X$ is finite.
Since $X$ is finite, $H$ is finite. Moreover, we can write
$$X = \{x_1, x_2, \dots, x_n\} \hspace{1cm} H = \{h_1, h_2, \dots, h_m\}$$
Since $f$ is injective, $h_j \neq h_k$ implies $f(h_j) \neq f(h_k)$. But recall that $H$ is proper, so $\exists \, x_i \in  X \setminus H$.

But at this point, I get stuck. Intuitively, I understand that this can not be a bijection, but I'm having trouble showing this.
I have already read Proof that a set is infinite if and only if it has an infinite proper subset and this post does not provide an actual proof answer or assumes that the set is countable.
 A: We want to prove that:
(1) X is equipotent to a proper subset H if and only if (2) X is an infinite set.
Note: X is equipotent to H $\begin{smallmatrix} def \\ \iff \\ \space \end{smallmatrix}$ there is a bijection between X and H $\begin{smallmatrix} def \\ \iff \\ \space \end{smallmatrix}$   |X|=|H| (they have the same cardinality)
H is a proper subset of X $\begin{smallmatrix} def \\ \iff \\ \space \end{smallmatrix}$ H $\subsetneq$ X.
Proof
(1)$\Rightarrow$(2) Assume by contradiction that X is finite. Since there exist a bijection $f:X\rightarrow H$ then by definition of cardinality $|X|=|H|$. But H is a subset of X which is proper so $\forall h \in H : h \in X$ but $\exists x \in X : x \notin H$. Hence H has at least one element less than X, so $|H|<|X|$ - contradiction.
(2)$\Rightarrow$(1) To prove this direction in (ZF) we need the condition "X is a Dedekind-infinite set" instead of just an infinite set. However, we can prove it in (ZF + ACw) applying the axiom of countable choice:



In wikipedia you can find the proof that every infinite set is Dedekind-infinite in (ZF + ACw).
A: I am going to assume that the definition of a finite set is a set equipotent to $\{1 , \cdots ,n\}$ for some natural number $n$, because that seems to be the one that is being used in the question.
Notation: We denote $[|1,n|]$ the set of integers between $1$ and $n$.
Proposition:
Let $n \in \mathbb{N}$, $A \subseteq [|1,n|]$ and $f: [|1,n|] \longrightarrow  A$ and injective map. Then $A=[|1,n|]$ and $f$ is a bijection.
We will prove this statement by induction on $n$.
For $n=0$, if $A$ is subset of an empty set, then $A$ has to be equal to the empty set and there is only one map from the empty set to itself and it is a bijection.
Now assume the statement true for $n\in \mathbb{N}$. Let us prove it is true for $n+1$. Take an injective map $f$ from $[|1,n+1|]$ to $A\subseteq [|1,n+1|]$.
By contradiction, assume that $k \not \in A$ for some $k \in [|1,n+1|]$. Let $\sigma$ denote the transposition $(k \ n+1)$, that is to say the permutation on $[|1,n+1|]$ that swaps $k$ and $n+1$ and fixes everything else. Denote $A'= \sigma (A)$. The map $\sigma \circ f$ restricted to $[|1,n|]$ then is an injective map from $[|1,n|]$ to $A'$, so by assumption $A'=[|1,n|]$ and $\sigma \circ f$ is a bijection from $[|1,n|]$ to $A'$.
That means that $\sigma \circ f(n+1)=n+1$, since $\sigma \circ f$ is injective. That means that $f(n+1)=k$ , which is a contradiction.
Now let us prove that $f$ is surjective. By contradiction if there exists a $k \in [|1,n+1|]$, such that $\forall x \in [|1,n+1|], f(x) \neq k$, then $f$ is an injective map from $[|1,n+1|]$ to $[|1,n+1|] \setminus \{k\}$, which we have already shown cannot be the case.
Proposition: If a set $X$ is equipotent to a proper subset, then it is not finite.
By contradiction assume that $X$ if finite and that there exists $Y \subseteq X$ a proper subset and $g$ a bijection from $X$ to $Y$. Let us denote $\phi$ a bijection from $X$ to $[|1,n|]$ for some $n \in \mathbb{N}$. The map $\phi \circ g \circ \phi^-1$ is then an injective map from $[|1,n|]$ to $\phi (Y) \subseteq [|1,n|]$, therefore by the previous proposition:
$\phi (Y)=[|1,n|]$. By applying $\phi^{-1}$ to both sides of this equation, we get that $Y=X$, which is a contradiction.
