Prove that problem is $NP$-complete: Given alphabet $A$ and regular expression check if this regular expression generate word containing all symbols in $A$.
Keep in mind that alphabet is not fixed. I have no idea how to solve it. The good thing it to prove $NP$-hard at begin I think. I would like to reduce CNF-SAT problem, however I can't start.
OP's idea is correct; $\sf{3\text{-}SAT}$ can be reduced to this problem. Given a CNF with $m$ clauses $(A,B,C,D,\ldots)$ and $n$ variables ($x_1,x_2,\ldots,x_n$), where each clause contains three literals, we can construct a corresponding regular expression on an alphabet with $m$ symbols; the regex will accept a word with all $m$ symbols if and only if the CNF is satisfiable. Specifically, the regex consists of $n$ blocks of the form $\sf{(a_i \;\vert \; b_i)}$, where $\sf{a_i}$ ($\sf{b_i}$) is a string of the symbols corresponding to exactly those clauses that contain $x_i$ ($\neg x_i$). (If $\sf{a_i}$ is empty, replace this by $\sf{(b_i)?}$, and vice-versa.) For example, if the CNF is $$ (x_1\vee\neg x_2\vee \neg x_3)\wedge(x_4\vee x_1\vee x_2)\wedge(\neg x_4 \vee x_5 \vee \neg x_2), $$ then the regular expression is $$ \sf{(AB)?(B\;\vert\;AC)A?(B\;\vert\;C)C?}. $$ Any assignment of truth variables corresponds to a path through the regular expression, making a binary choice for each variable (resp., each block); a satisfying assignment (word) is one that makes every clause true (includes every symbol).