Does *finitely many* include the option *none*?

Yes. No question. A subset of a finite set is finite. A polynomial with real coefficients has finitely many real zeros. We do not need (or want) to require saying: "A subset of a finite set is either finite or empty".

Gerd Faltings showed that $a^{n}+b^{n}=c^{n}$ has finitely many solutions $(a,b,c)$ for any $n>2$. 10 years later, Andrew Wiles showed that it had none. Faltings is not known to have contradicted.