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A formal language with finitely many words. Every finite language is a regular language: take a finite union of the regular expressions spelling its members, with the empty language represented separately when there are no members.

Regular languages are closed under finite intersection, but not countably infinite intersection. Indeed, for any formal language $K\subseteq {\Sigma }^{\ast }$ over a finite alphabet, let $L_i$ contain every word of length at least $i$, together with the members of $K$ of length less than $i$. Each $L_i$ is a regular language, since its exceptional short part is a finite language, and ${\bigcap }_{i\ge 1}{L}_i=K$. Thus every formal language, including one that is not a regular language, can arise in this way.

No: regular languages are not closed under countably infinite intersection. Let $\Sigma =\{0,1\}$ and putThe first term in $L_i$ has regular expression $(0\cup 1{)}^i(0\cup 1{)}^{\ast }$, where the fixed power $i$ means $i$ copies. The second term is a finite language, a finite union of single words. Their union is therefore a regular language.

If $w\in K$, then for each $i$ either $|w|\ge i$ or $w$ belongs to the second term, so $w\in L_i$. If $w\notin K$, choose $i>|w|$. It belongs to neither term of $L_i$. ConsequentlyThe formal language $K$ is not a regular language. Indeed, for a proposed pumping length $p\ge 1$, the pumping lemma for regular languages applied to $0^p{1}^p$ must pump a nonempty block consisting only of zeros. Repeating that block increases the number of zeros without increasing the number of ones, leaving $K$. This contradiction proves the counterexample. Finite intersection closure does not extend to this infinite intersection of regular languages.