Surjection between two finite sets of different sizes

Recalling the definition of function, two elements from $B$ can't correspond to the same element in $A$. (Considering $f:A\to B$).

Since you have $|B|=4>2=|A|$, the definition of suerjection won't be satisfied.

A fully rigorous argument here can be extremely hard. To give a mostly rigorous argument: Suppose for contradiction that $f:A\to B$ is a surjection. Then $f(A)⊆ B$ and $f(A)=\{f(1),f(2)\}$ so that there are only two elements in $f(A)$. Since there are four elements in $B$ then there must be some element in $B\smallsetminus f(A)$ (the "relative complement"). If we call this element $y$ then there is no $x$ such that $f(x)=y$.