Prove or Disprove: an infinite intersection of regular languages is a context-free language
&quot;I know that this statement is false, but couldn't find an example to disprove it.&quot;
It might come as a surprise to you that, in fact, every non-context-free language can be a counterexample. We have the following fact, assuming any fixed alphabet $\Sigma$.

Let $L$ be a language. Then there exist regular languages $L_1, L_2,...$ such that $L=\cap_{k=1}^{\infty}L_k$.
Proof. The other answer indicates how to construct $L_i$. For clarity, here is the construction.

Let $L_i$ be the words in $L$ with length at most $i$ together with all the words with length greater than $i$. More formally, let $L_i=\{w\in\Sigma^*: w\in L\text{ and } |w|\le i\}\cup\{w\in\Sigma^*:|w|\gt i\}$. Then $L=\cap_{k=1}^{\infty}L_k$.
Each $L_i$ is regular since $L_i$ is the union of two regular languages.

$\{w\in\Sigma^*: w\in L\text{ and } |w|\le i\}$, as a finite set, is regular.
$\{w\in\Sigma^*:|w|\gt i\}$ is regular.


So, even if $\forall k\in \mathbb{N} $, $L_k$ is a regular language, $L=\cap_{k=1}^{\infty}L_k$ can be non-regular, non-context-free, non-context-sensitive, etc.