I am going to replace some definitions by obviously equivalent ones. The benefit will be that the proofs are easier to understand. I am writing down very detailed proofs that you would not normally find in a college-level math textbook.
Definition: For $n \in \mathbb{N}$ define $[n] = \{k \in \mathbb{N} \mid k < n\}$.
Definition: The image of a map $f : X \to Y$ is the set $\mathrm{im}(f) = \{y \in Y \mid \exists x \in X \,.\, f(x) = y\}$.
Definition: A set $X$ is infinite when for every $n \in \mathbb{N}$ and every map $f : [n] \to X$ there is some $x \in X$ which is not in the image of $f$.
Lemma 1: An infinite set $X$ contains an element.
Proof. Because $X$ is infinite there is $x \in X$ which is not in the image of the unique map $[0] \to X$. QED.
Lemma 2: If $X$ is infinite then there exists a map $c$ such that $c(n,f) \in X \setminus \mathrm{im}(f)$ for every $n \in \mathbb{N}$ and every $f : [n] \to X$.
Proof. For every $n \in \mathbb{N}$ and $f : [n] \to X$ the set $X \setminus \mathrm{im}(f)$ contains an element because $X$ is infinite. By the axiom of choice there exists a choice map $c$ such that $c(n,f) \in X \setminus \mathrm{im}(f)$ for all $n$ and $f$. QED.
Here is an informal explanation of the map $c$. Given an infinite set $X$ and any elements $x_0, \ldots, x_{n-1} \in X$, we may view the tuple $(x_0, \ldots, x_n)$ as a map $f : [n] \to X$ where $f(k) = x_k$. By a slight abuse of notation we can then write $c(n,f)$ as $c(x_0, \ldots, x_{n-1})$. Now we see that $c$ accepts finitely many elements of $X$ and returns an element in $X$ which is different from all of them.
Proposition 1: If $X$ is infinite then there exists an injective map $i : \mathbb{N} \to X$.
Proof. By Lemma 1 there is $x \in X$ and let $c$ be the map from Lemma 2. Define $i$ by recursion:
\begin{align*}
i(0) &= x \\
i(n) &= c(i(0), \ldots, i(n-1)) \qquad\qquad \text{for $n \geq 1$}
\end{align*}
Clearly, $i$ is injective because $i(n)$ is chosen so that it is different from $i(0), i(1), \ldots, i(n-1)$. QED.
Proposition 2: If $X$ is infinite then it is in bijection with some proper subset $Y \subseteq X$.
Proof. Suppose $X$ is infinite. We seek a proper subset $Y \subseteq X$ and a bijection $f : X \to Y$. Let $i$ be the map from Proposition 1. Define
$$Y = X \setminus \{i(0)\}$$
and define $b : X \to Y$ by
$$b(x) = \begin{cases}
x & \text{if $x \not\in \mathrm{im}(i)$}\\
i(n+1) & \text{if $x = i(n)$ for a (unique) $n \in \mathbb{N}$}
\end{cases}
$$
in words, $b$ takes $i(0)$ to $i(1)$, $i(1)$ to $i(2)$, and so on, and does not move elements which are outside the image of $i$. Clearly, $b$ is injective because $i$ is. It is surjective because the only element not in the image of $b$ is $i(0)$. QED.
Proposition 3: If $X$ is a set and $b : X \to Y$ a bijection from $X$ to a proper subset $Y \subseteq X$, then there is an injective map $j : \mathbb{N} \to X$.
Proof.
Because $Y$ is a proper subset there exists $x \in X \setminus Y$. We define a map $j : \mathbb{N} \to X$ by recursion:
\begin{align*}
j(0) &= x \\
j(n) &= b(j(n-1)) \qquad\qquad\text{for $n \geq 1$}
\end{align*}
To prove that $j$ is injective, it suffices to verify that $j(m) \neq j(n)$ for all $m < n$. We do this by induction on $m$:
if $m = 0$ then $j(0) = x$ is different from $j(n) = b(j(n-1))$ because $b(j(n-1)) \in Y$ and $x \not\in Y$.
for the induction step, suppose we had $j(m) = j(n)$ for some $0 < m < n$. Then we would have $b(j(m-1)) = j(m) = j(n) = b(j(n-1))$ and because $b$ is a bijection it would follow that $j(m-1) = j(n-1)$, which is impossible by the induction hypothesis for $m-1$. Therefore it must be the case that $j(m) \neq j(n)$.
QED.
Theorem: A set is infinite if, and only if, it is equipotent with some proper subset.
Proof.
If $X$ is infinite then we apply Propostion 2 to get a proper subset $Y \subseteq X$ which is equipotent with $X$.
Conversely, suppose we have a bijection $b : X \to Y$ for a proper subset $Y \subseteq X$. By Proposition 3 there is an injective map $j : \mathbb{N} \to X$. To see that $X$ is infinite, consider any $n \in \mathbb{N}$ and $f : [n] \to X$. There are $n+1$ distinct elements $j(0), j(1), \ldots, j(n) \in X$, and so at least one of them is not in $\mathrm{im}(f)$ because $\mathrm{im}(f)$ has at most $n$ elements. QED.
Best Answer
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).