We know that if there are two languages, $L_1$ and $L_2$, if $L_1$ and $L_2$ are regular, the intersection of those two is also regular. Suppose we have two machines, $M_1$ and $M_2$, and using them, a new machine $M_3$ is built. If $M_1$ had $k>0$ states and $M_2$ had $l>0$ states, the new machine has to have $kl$ states at minimum.
How would I go about trying to prove this? I've been stuck on this for quite some time. I know that we would have to construct the languages $L_1$ and $L_2$ so that $M_1$ and $M_2$ accept them, respectively, which in turn allows the smallest machine accepts $L_1$ intersects $L_2$ has $k \times l$ states.
Any help would be greatly appreciated.
I'm wondering whether maybe you got the question the wrong way round, and what you're actually supposed to show is that you need at most $kl$ states. Because that is true, whereas what you have asked about is not, as Qiaochu already mentioned in the comments (if $L_1$ and $L_2$ are the same language, then $L_1\cap L_2 = L_i$ for $i=1,2$ and so you can use either $M_1$ or $M_2$ to recognise the intersection, so you only need $\min\{k,l\}$ states in that case).
Let me know if you'd like an explanation of why you need at most $kl$ states, but you might find you can do the question once you realise what it actually says.