How can the intersection of CFLs and REGs be CFL if REG is a proper subset of CFL?
You are confusing two different statements.


$\mathrm{CFL} \cap \mathrm{REG} = \mathrm{REG}$ or, equivalently,
 for all $L_1 \in \mathrm{REG}$ : $L_1 \in \mathrm{CFL}$.
For all $L_1 \in \mathrm{REG}$, $L_2 \in \mathrm{CFL}$ : $L_1 \cap L_2 \in \mathrm{CFL}$.


The first makes a statement on two sets of languages (aka language classes), the second makes one over (many) pairs of sets of strings (aka languages). So in mixing the two, you are creating a type error.

Also, you seem to suggest that if a language is context-free, it's not regular. That is false; we have $\mathrm{REG} \subsetneq \mathrm{CFL}$ (cf. 1.), so showing a language is CFL (using 2.) does not preclude it from being regular. Specifically, it's easy to find examples of a non-regular context-free and a regular language whose intersection is regular.