Does this proof work for infinite regular languages My proof was deemed false because it does not work for infinite regular languages, but I don't understand why.

Prove: "If we remove one string from any nonempty regular set, the resulting set is still regular.

I used to lemmas:


A set containing only one string is trivially a regular language
The union operator is closed in the class of regular languages


Given these two lemmas, I made the claim that any regular language $L$ can be written in the form
$$L = \bigcup_{e \in L}\{e\}$$

Then, one picks an arbitrary string $r \in L$ to remove, and expresses $L$ in the same manner with $r$ distinct:
$$L = \bigcup_{e \in (L - \{r\})}\{e\} \cup \{r\}, \quad r \in L$$

Finally, we simply remove $r$ from the union and obtain:
$$L' = \bigcup_{e \in (L - \{r\})}\{e\}$$

$L'$ is also regular by the two lemmas introduced earlier.
The problem with your argument is in the interpretation of the second lemma. Closure under union says "If $L_1$ and $L_2$ are regular languages, then $L_1\cup L_2$ is regular." By induction, then, the union of any finite collection of regular languages is regular. You cannot, however, conclude that the union of an arbitrary collection of regular languages is regular, as Raphael pointed out, since your observation
$$
L=\bigcup_{e\,\in\, L}\{e\}
$$
is true for any language at all, regular or not, so since a single string language is regular, then every language would be regular.



What you could do is exploit some other closure properties of regular languages. For instance


If $L_1$ and $L_2$ are regular languages, so is $L_1\cap L_2$.
If $L$ is a regular language, its complement $\overline{L}$ is too.
We can thus write the set difference $A\setminus B$ as $A\cap\overline{B}$ so regular languages are closed under set difference.
If $L$ is any regular language and $e\in L$ is any string in $L$, then $L\setminus \{e\}$ must also be regular.