= Solution
\b[No: <regular languages> are not closed under countably infinite <intersection>.] Let $\Sigma=\{0,1\}$ and put
$$
K=\{0^n1^n:n\geq0\},\qquad
L_i=\{w\in\Sigma^*:|w|\geq i\}\cup\{0^n1^n:2n<i\},\quad i\geq1.
$$
The first term in $L_i$ has <regular expression> $(0\cup1)^i(0\cup1)^*$, 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 $|w|\geq i$ 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 $L_i$. Consequently
$$
\boxed{\bigcap_{i=1}^{\infty}L_i=K.}
$$
The <formal language> $K$ is not a <regular language>. Indeed, for a proposed <pumping length> $p\geq1$, the <pumping lemma for regular languages> applied to $0^p1^p$ 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>.
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