Proving a language comprised of 2 languages is regular
Proof using closure properties

Let $add\colon \Sigma \to \Sigma^*$ be a regular substitution that maps $\mu \in \Sigma$ to $\{ \sigma \mu \xi : \sigma, \xi \in \Sigma \}$ (the image is regular since it is finite). Then your language is $L_1 \cap add(L_2)$.

Proof using product automaton

Let $(\Sigma, Q^i, q_0^i, F^i, \delta^i)$ be automata accepting $L_i$ (for $i = 1,2$). We construct a new automaton $(\Sigma, Q, q_0, F, \delta)$ where:


$Q = Q^1 \times Q^2 \times \{1,2,3\}$.
$q_0 = (q_0^1,q_0^2,1)$.
$F = F_1 \times F_2 \times \{1\}$.
$\delta((q^1,q^2,1),\sigma) = (\delta^1(q^1,\sigma),q^2,2)$.
$\delta((q^1,q^2,2),\sigma) = (\delta^1(q^1,\sigma),\delta^2(q^2,\sigma),3)$.
$\delta((q^1,q^2,3),\sigma) = (\delta^1(q^1,\sigma),q^2,1)$.


The product automaton keeps track of the state in both automata, as well as the location within the triple $\sigma_i \mu_i \xi_i$ (this is the purpose of the third argument).