The alternating group $A_4$ is generated by {(12)(34),(123)}. Let $a = (12)(34)$, $b = (123)$. We see that $a^2 = 1, b^3 = 1$:
$$No.1:a^0 b^0 = 1$$
$$No.2: a^1 b^0=(12)(34)$$
$$No.3: a^0 b^1=(123)$$
$$No.4:b^2 = (123)(123) = (132)$$
$$No.5:ab = (12)(34)(123) = (243)$$
$$No.6:ab^2 = (12)(34)(123)(123) = (143)$$
$$No.7:ba = (123)(12)(34) = (134)$$
$$No.8:b^2a = (123)(123)(12)(34) = (123)(134) = (234)$$
$$No.9:bab = (123)(12)(34)(123) = (134)(123) = (124)$$
$$No.10:bab^2 = (123)(12)(34)(123)(123) = (124)(123) = (14)(32)$$
$$No.11:b^2 a b = (123)(123)(12)(34)(123) = (234)(123)=(13)(24)$$
$$No.12:b^2 a b^2 = (123)(123)(12)(34)(123)(123) = (123)(14)(32) = (142)$$
But 4 elements has 24 permutations, so half of them shall be in $|A_4|$.
Since those 12 are distinct, done.
Best Answer
A(nother) conceptual proof goes like that.
$a = (12)(34)$ and $b = (123)$ are even permutations, so $H = \langle a, b \rangle \le A_{4}$.
$H$ contains an element of order $2$ and an element of order $3$, so it has order divisible by $2$ and $3$, hence its order is a multiple of $6$.
Now $A_{4}$ does not have a subgroup $H$ of order $6$. Such a subgroup would have index $2$ in $A_{4}$, and thus it would be normal in $A_{4}$. Now $H$ contains a normal subgroup of order $3$, and thus $A_{4}$ would have a unique subgroup of order $3$, which is clearly not the case.