Decidability of CFG that accepts at least one odd length string Let $L$ be the language$\{\langle G \rangle : G\text{ is a CFG and L(G) accepts at least one odd-length strings}\}$.
  
  Prove that $L$ is decidable or undecidable.


Here is my attempt:

$L$ = On input $\langle G \rangle$, where $\langle G \rangle$ is a CFG:


Construct CFG, such that $L(O)$ contains only odd length strings.
Construct CFG $\langle H \rangle$ such that $L(H) = L(O) \cap L(G)$ 
Input $\langle H \rangle$ to the decider $E_{\text{cfg}}$
If it accepts, $reject$.
Else, $accept$. 


Is my proof correct? In the first step, can I construct a DFA, $O$ such that $L(O)$ consists of only odd length strings instead of a CFG?
You cannot compute the intersection of two context-free grammars, at least not if you expect the answer to be another context-free grammar. The reason is that context-free languages are not closed under intersection (consider $a^nb^nc^m \cap a^mb^nc^n = a^nb^nc^n$).

However, context-free languages are closed under intersection with a regular language, and moreover there is an efficient algorithm that, given a context-free grammar $G$ and an NFA $A$, computes a new context-free grammar for the language $L(G) \cap L(A)$. You should use this algorithm in your second step.

Another problem with your solution is that it's not clear what you mean by "$L(O)$ contains only odd length strings". There are many languages that contain only odd length strings, for example the empty language. You want $L(O)$ to contain only and all odd length strings, that is, you want it to be the language of odd length strings.