Please check my below proof. My proof is somewhere messy since I don't know how to organize and present ideas efficiently. I'm happy to receive any suggestion to have a shorter, more concise, and more elegant proof 🙂
Theorem:
Suppose that $A$ is finite and that $f:A \to B$ is surjective. Then $B$ is finite and $\vert{B}\vert \leq \vert{A}\vert$, the equality holds $\iff f$ is bijective.
Proof:
$A$ is finite $\implies$ there exists a bijection $t:I_n \to A$ where $I_n$ is an initial segment of $\mathbb{N}$.
$\implies f\circ t:I_n \to B$ is a surjection.
Let $g:B \to I_n$ s.t $g(b)=\min(f\circ t)^{-1} \{b\}$.
If $g(b)=g(c)$, then $\min(f\circ t)^{-1} \{b\}=\min(f\circ t)^{-1} \{c\}$. This implies there exists $m$ s.t $f\circ t (m)=b$ and $f\circ t (m)=c$. Thus $b=c$. So $g$ is injective.
$\implies g:B \to g[B]$ is bijective. $g[B] \subseteq I_n \implies$ g[B] is finite.
As a result, $B$ is finite.
$g[B] \subseteq I_n \implies \vert{B}\vert \leq \vert{I_n}\vert=\vert{A}\vert \implies \vert{B}\vert \leq \vert{A}\vert$. The equality holds $\iff g[B] = I_n \iff g:B \to I_n$ is bijective.
Now we prove ($g:B \to I_n$ is bijective) $\iff (f$ is bijective). It is easy to show that $(g:B \to I_n$ is bijective) $\Leftarrow (f$ is bijective). So our task is to prove $(g:B \to I_n$ is bijective) $\implies (f$ is bijective).
Assume that $f(a_1)=f(a_2)=b$. Then $\exists x_1,x_2 \in I_n$ s.t $f \circ t(x_1)=f \circ t(x_2)=b \implies x_1,x_2 \in \{m \in \mathbb{N} \mid f\circ t (m)=b\}$.
Assume $x_1 \neq x_2$. Without loss for generality, we assume $x_1 < x_2$. This implies $x_2 \neq \min(f\circ t)^{-1} \{b\} \implies \not \exists b \in B$ s.t $g(b)=x_2 \implies g$ is not surjective (CONTRADICTION).
Thus $x_1=x_2$ or equivelently $a_1=a_2$.
To sum up, $f(a_1)=f(a_2)=b \implies a_1=a_2$. As a result, $f$ is injective $\implies f$ is bijective.
Best Answer
I found that my previous proof is ambiguous at important arguments. So I decided to rewrite it here.
Let $I_n=\{i\in\Bbb N\mid i\le n\}$. Since $A$ is finite, there is a bijection $g:I_n\to A$. Thus $f\circ g: I_n\to B$ is surjective. We define a mapping $h:B\to I_n$ by $h(b)=\min\{i\in\Bbb N\mid f\circ g(i)=b\}$.
For $b_1,b_2\in B$, $h(b_1)=h(b_2) \implies \min\{i\in\Bbb N\mid f\circ g(i)=b_1\}=\min\{i\in\Bbb N\mid f\circ g(i)=b_2\}=\bar i$ $\implies f\circ g(\bar i)=b_1$ and $f\circ g(\bar i)=b_2 \implies b_1=b_2$. Thus $h$ is injective and consequently $h:B\to h[B]$ is bijective. Moreover, $h[B] \subseteq I_n$ and $I_n$ is finite. Hence $B$ is finite. We have $|B|=|h[B]|\le |I_n|=|A|$, then $|B|\le |A|$.
The equality holds $\iff |B| = |A| \iff h[B]=I_n \iff h:B \to I_n$ is bijective. So our task is to prove $h$ is bijective $\iff f$ is bijective. As $h$ is already injective and $f$ is already surjective, our task is to prove $h$ is surjective $\iff f$ is injective.
a. $h$ is surjective $\implies f$ is injective
For $a_1,a_2\in A$ and $f(a_1)=f(a_2)=b$. Since $a_1,a_2\in A$, there exist $i_1,i_2\in I_n$ such that $g(i_1)=a_1$ and $g(i_2)=a_2$. Then $f\circ g(i_1)=f\circ g(i_2)=b$. Then $i_1,i_2\in \{i\in\Bbb N\mid f\circ g(i)=b\}$. Assume $i_2>i_1$, then $i_2>i_1\ge \min\{i\in\Bbb N\mid f\circ g(i)=b\} =h(b)$ and consequently $i_2 \neq h(b)$. $f\circ g(i_2)=b\neq b'$ for all $b'\in B$ and $b'\neq b$. Then $i_2 \notin \{i\in\Bbb N\mid f\circ g(i)=b'\}$ for all $b'\in B$ and $b'\neq b$. Then $i_2 \neq \min\{i\in\Bbb N\mid f\circ g(i)=b'\}$ for all $b'\in B$ and $b'\neq b$. Then $i_2 \neq h(b')$ for all $b'\in B$ and $b'\neq b$. To sum up, $i_2 \neq h(b')$ for all $b'\in B$. Thus $i_2 \notin \operatorname{ran}h$. This contradicts the surjectivity of $h$. It follows that $i_1=i_2$, then $g(i_1)=g(i_2)$, then $a_1=a_2$. Hence $f$ is injective.
b. $f$ is injective $\implies h$ is surjective
For $m\in I_n$, if $f\circ g(k)=f\circ g(m)$, then $g(k)=g(m)$ by the fact that $f$ is injective. Then $k=m$ by the fact that $g$ is bijective. Hence $m=\min\{i\in\Bbb N\mid f\circ g(i)=f\circ g(m)\}$. Thus $h(f\circ g(m))=m$ where $f\circ g(m)\in B$. It follows that $h$ is surjective.