Are regular languages closed by a full-shuffle operation? 
Let be two languages $L_1,L_2$ we define the operation full-shuffle as $S(L_1,L_2)=\{w \mid w=w_1w_2...w_k\}$ such that $w_1,w_3,...\in L_1$ and $w_2,w_4...\in L_2$. In other words, the language $L$ only contains words that we can build from a word $L_1$ followed by a word from $L_2$, followed by a word from $L_1$, followed by a word from $L_2$, and so on...

How can I show that regular languages are closed by this operation ?
I have some difficulties to imagine the formalize automata resulting from this.
I did the following from p49, in Introduction to theory of computation from A. Meheshwari and M. Smid.
Let $M_1=(Q_1,\Sigma,\delta_1,q_1,F_1)$ be an NFA such that $A_1=L(M_1)$ and $M_2=(Q_2,\Sigma,\delta_2,q_2,F_2)$ be an NFA such that $A_2=L(M_2)$. We will construct an NFA $M=(Q,\Sigma,\delta,q_0,F)$ such that $L(M)=A_1A_2A_1A_2A_1...$


*

*$Q = Q_1 \cup Q_2$ ($\cup Q_1 \cup Q_2...?$ but it remains the same isn't it ?)

*$q_0=q_1$

*$F=F_1 \cap F_2$ as far as we have to have words from both $L_1$ and $L_2$.

*$\delta : Q\times\Sigma_\epsilon \rightarrow P(Q)$ is defined as folows :


$$\delta(r,a)=\begin{cases}
\delta_1(r,a)\mbox{ if $r\in Q_1$}\\
\delta_2(r,a)\mbox{ if $r\in Q_2$}\\
\end{cases}$$
I'm new to this kind of proof and I'm not sure about it.

Just use regular expressions instead of automata. Indeed,
$$S(L_1,L_2) = (L_1L_2)^* \cup (L_1L_2)^*L_1$$
