I know there are no elements of $S_4$ with order 12 from a list of the elements of $S_4$ but how can I prove it without listing all the elements with their orders?
Every permutation can be written as a product of disjoint cycles. The order is the least common multiplier of the cycle-lengths.
If we have $4$ elements, order $12$ would require a $3$-cycle, but then there is no place for another cycle.
The possible orders for permutations with $4$ elements are $1,2,3,4$ because in the case of two non-trivial cycles, we can only have two transpositions.