Proof that finite automata is closed under intersection I'm looking at a proof that says that:
If $M_1=(Q_1, \Sigma , q_1, A_1, \delta)$ and $M_2=(Q_2, \Sigma , q_2, A_2, \delta)$ are two finite automata(FA) then $M=M_1 \cup M_2$ is also an FA. We define $M=(Q, \Sigma , q, A, \delta)$ where $Q=Q_1 \times Q_2$, $q_0=(q_1,q_2)$ and $\delta((p,q),\sigma)=(\delta(p,\sigma),\delta(q,\sigma))$.

I'm just having one problem. We have to prove that $\delta^*((p,q),x)=(\delta^*(p,x),\delta^*(q,x))$. (x is a string in $\Sigma^*$)
We do it by induction. I've already proven the base case, and I'm now trying to prove it holds for $x'=x\sigma$. This is what I've gotten so far:

$\delta^*((p,q),x')=\delta(\delta^*((p,q),x),\sigma)=\delta(\delta^*(p,x),\delta^*(q,x)),\sigma)$

I don't really know how to proceed from here. Can someone help?
A possible source of your confusion is that the transition functions of $M_1$, $M_2$ and $M$ are different functions, yet you call all of them $\delta$. If instead you use $\delta_1$ resp. $\delta_2$ for the transition function of $M_1$ resp. $M_2$, the definition of $\delta$ changes to $\delta((p,q),\sigma)=(\delta_1(p,\sigma),\delta_2(q,\sigma))$. You should be able to add the subscripts in the appropriate positions of your partial proof yourself.

Now, after noting that $\delta_1^*(p,x)\in Q_1$ and $\delta_2^*(q,x)\in Q_2$, the proof can be completed by a straightforward application of the definitions of $\delta$ and $\delta_i^*$.