Let $G=(V,T,P,S)$ be a context-free grammar without $\epsilon$ rules. Define a context-free grammar $G'$ which creates a language which consists of all words from $L(G)$ without one of the letters of a word which belongs to $L(G)$. For example, if $ab\in L(G)$ then $a\in L(G')$ and $b\in L(G')$.
The solution to the problem is as follows:
Let $G'=(V\cup V', T,P',S')$. $$ \forall (A\to \alpha)\in P\implies (A\to \alpha)\in P'\\ \forall (A\to \alpha B\gamma)\in P, B\in V\implies (A\to \alpha B\gamma)\in P'\\ \forall (A\to \alpha t\gamma)\in P, t\in T\implies (A\to \alpha\gamma)\in P' $$
Why $\forall (A\to \alpha B\gamma)\in P'$, because $\alpha, \gamma$ are strings of stack symbols (not just letters) hence they cannot be divided (like for example, $\forall (A\to \alpha B\gamma)\in P, B\in V\implies (A\to \alpha)\in P'\quad\land\quad (A\to \gamma)\in P'$)?