Closure under the Star Operation
One must be careful about what the closure properties actually state.

If $A,B$ are regular languages, then $AB$ and $A \cup B$ are regular languages as well.

By induction, one can obtain that the concatenation / union of finitely many regular languages $A_1,\ldots,A_n$ is regular.

This however does not extend to infinite unions!

To understand why the extension would be wrong, consider a simpler case, first. Take the class of finite languages: sets containing finitely many words. Clearly this class is closed under union: if $A,B$ are finite languages, then $A\cup B$ is a finite language. However, taking the union of the infinitely many finite languages $\{1\},\{11\},\{111\},\ldots$ one gets an infinite language $\{1,11,111,\ldots\}$.

In the realm of regular languages, we met the same problem. Take any non-regular language $L$: it has to be infinite, since otherwise it would be regular. We can write $L=\{w_1,w_2,\ldots\}$ for some $w_1,w_2,\ldots$. But we also have $L=\{w_1\}\cup\{w_2\}\cup\ldots = \bigcup_{i\geq 1}\{w_i\}$, which is an infinite union of singleton-languages, which are regular. So, any non regular $L$ can be written as an infinite union of regular languages.

We conclude that the union of infinitely many regular languages is not necessarily a regular language.