We have $23 \times 46=1058$. We will prove that $\overline{EFGH} \ge 1058$ then. Let $A<C$, then if $A \ge 3$ we have $\overline{AB} \times \overline{CD} \ge 30 \times 40>1058$. So $A=2$. If $C \ge 5$ then $\overline{AB} \times \overline{CD} \ge 23 \times 54>1058$ (as $A,B,C,D,E,F,G,H$ should be different digits and we need $E=1,F=0$ if $\overline{EFGH}<1058$). So $C=3$ or $4$. If $C=3$, we can try all the values of $A,B,C,D$ and we can find that $\overline{EFGH} \ge 1058$. If $C=4$, try from $\overline{AB}=23$ then we have $1058$ is the minimum possible value of $\overline{EFGH}$. if $\overline{AB} \ge 25$ then $\overline{AB} \times \overline{CD} \ge 25 \times 43>1058$. btw, isn't it a primary school problem?

No, it appeared on the most recent Irish Mathematical Olympiad on Paper 2. Nice solution anyways.