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Below I present a proof which is probably an overkill, however whenever I tried to simplify things I ended up finding it necessary to use some relatively deep stuff.

If anyone can provide a more elementary proof, I'll either delete this answer or add the proof to the answer, depending on how easy it is.

First a definition which I didn't explicitly use, but it's useful for whoever tries to give a more elementary proof.

Definition: A set is said to be finite if it is equinumerous to (with?) $[k\textbf{]}\color{grey}{(=\{x\in \omega :x<k\})}$, for some $k\in \omega$.

Statement: Let $A,B$ be finite sets. If $|B|\leq |A| \land A\subseteq B$, then $A=B$.

Proof: Suppose that $|B|\leq |A| \land A\subseteq B$ and $A

eq B$. Then $A\subset B$ and since they are finite $|A|<|B|$, (this uses the fact that a set is finite if, and only if, it isn't Dedekind infinite which is something way too deep considering out strongly the statement is assessed as intuitively true). And this is a contradiction due to the trichotomy of cardinal numbers (which isn't skin deep either).

The following proof was suggested (and typed) by T. Verron. Even though it is elementary it still serves my purpose: the gap between how easy it is to believe the statement and the proof is huge.

A possible elementary proof: Let $m = |A|$ and $n= |B|$. Let $f$ be a bijection from $A$ to $[m]$, and let $g$ be a bijection from $B$ to $[n]$. Define a new map $h : B \to [n]$, whose restriction to $A$ is $f$. For example, let $\sigma$ be a permutation of $B$, such that $\sigma(g^{-1}[m]) = f^{-1}([m])=A$, and define $h$ by

$h(x) = \begin{cases}f(x) & \text{if}\;\; x \in A \\ g \circ \sigma^{-1}(x) & \text{else}\end{cases}$

Then $h$ is a bijection from $B$ to $[n]$, and $h \circ f^{-1}$ is an injection from $[m]$ to $[n]$. That shows that $m \leq n$, and thus by assumption, that $m = n$. Now $f^{-1} \circ h$ is a bijection from $A$ to $B$, and by construction is the identity.