proof that a language is a regular according to another language let $L\subseteq\Sigma^*$ be a regular language.
for $\sigma \in \Sigma$ prove that $L'=\{w_1\sigma w_2:w_1w_2\in L\}$ is a regular languge.
I tried induction on the length of the regular expression of L.
there are 3 cases in the induction:
$r=r_1\cup r_2, r =r_1\cdot r_2 , r=(r_1)^*$.
the first one is pretty easy but now I stuck on the 2nd and the 3rd.
in the 2nd case: 
$w_1\sigma w_2\in L' \leftrightarrow w_1w_2\in L=L(r)=L(r_1\cdot r_2 )=L(r_1)\cdot L(r_2 )$
but at this point I stuck cause I can't say that $w_1\in L(r_1)$  or something like that.
in the 3rd case:
$w_1\sigma w_2\in L' \leftrightarrow w_1w_2\in L=L(r)=L((r_1)^*)=L(r_1)^* \leftrightarrow w_1 w_2=\epsilon$  or  $w_1w_2=w_1\cdot w_2...w_k, k>0, \forall i =1...k, w_i \in L(r_1)$
and I have no idea how to finish that case.
 A: The problem can be solved using automata, but your suggested approach is very interesting. However, you have to be very careful since there is nothing like "the regular expression of $L$" (a regular language admits infinitely many regular expressions that represent it). To avoid this potential problem, you can argue directly on regular languages as follows.
Let $a$ be a fixed letter of the alphabet $A$. For each language $L$ of $A^*$, let
$$
L' = \{uav \mid u,v \in A^*,\ uv \in L\}
$$
Let $\mathcal{C}$ be the class of all regular languages $L$ of $A^*$ such that $L'$ is regular. Our aim is to show that $\mathcal{C}$ contains all regular languages.
Step 1. The class $\mathcal{C}$ contains the empty language, the language $\{1\}$ and the languages $\{b\}$ for each letter $b$. Trivial (actually you could say directly that any finite language belongs to $\mathcal{C}$).
Step 2. The class $\mathcal{C}$ is closed under finite union. This follows from the formula $(L_1 \cup L_2)' = L_1' \cup L_2'$, which you already observed.
Step 3. The class $\mathcal{C}$ is closed under product. This follows from the formula $(L_1L_2)' = L_1L_2' \cup L_1'L_2$, which essentially says that if you want to insert an $a$ in $L_1L_2$, you have to insert it either in a word of $L_1$ or in a word of $L_2$.
Step 4. The class $\mathcal{C}$ is closed under star. This follows from the formula $(L^*)' = L^*L'L^* \cup \{a\}$ that I let you verify.
Thus $\mathcal{C}$ contains the finite languages and is closed under union, product and star: thus it contains all regular languages.
