Does this argument prove CFLs are not closed under union?
Your first proof is valid. You are assuming by contradiction that CFLs are closed under complement and arriving at the absurd conclusion that they must be closed under intersection. Notice that this works for any intersection operation between two CFL languages.

Your second proof is wrong. You would need to assume that CFLs are closed under union and arrive at a contradiction. In your case your are not showing that, under the assumption, CFLs are closed under complement and intersection (hence the absurd). You are just showing that CFLs must be closed w.r.t. a particular operation in which you perform complements and intersections in some particular order. 

Using the same logic I could prove the following erroneous claim.

Claim: CFLs are not closed under the identity operation $\textrm{Id}(L) = L$. 

"Proof": CFLs are not closed under complement and, given a CFL $L$, $\textrm{Id}(L)$ can be written as $\textrm{Id}(L) = L = \overline{\overline{L}}$.