Intersection of regular and not regular
To expand a bit on David's answer, you are confusing between the next two claims.

Claim 1 if $L_1,L_2$ are regular, then $L_1 \cap L_2$ is also regular.
(proof: construct the product DFA, e.g., here)

Claim 2 if $L_1$ is regular, but $L_2$ is not, then $L_1\cap L_2$ can be either regular or not-regular.
(proof: Let $L_2$ be some non-regular language. If $L_1=\Sigma^*$ then $L_1\cap L_2=L_2$ which is non-regular by its definition. On the other hand, if $L_1=\emptyset$ then $L_1\cap L_2=L_1=\emptyset$ which is regular)