Here is an incorrect proof that $E{Q}_{CFG}$ is decidable. $E{Q}_{CFG}$ is in fact not decidable.

Not a proof: Construct a TM as follows:

On input $($:$G_1,G_2$:$)$ :

Since CFLs are closed under complement, intersection, and union, we can construct a CFG $G$ for $(\frac{\text{:}}{b}ar(L(G_2))\}$

Run $M_{E_{CFG}}$ on $G$ and return the result.

This TM accepts inputs $($:$G_1,G_2$:$)$ with $L(G_1)=L(G_2)$ and rejects inputs $($:$G_1,G_2$:$)$ with $L(G_1)L(G_2),$ so it decides $E{Q}_{CFG}.$

Which of the following are issues with the proof?

A$.E_{CFG}$ is not decidable so $M_{E_{CFG}}$ does not halt on all inputs.

B$.$ CFLs are not closed under union, so we can't necessarily construct $G.$

C$.$ CFLs are not closed under intersection, so we can't necessarily construct $G$