If $R$ is an integral domain with unity having only finitely many subdomains (not necessarily with unity), then is it true that $R$ is finite ?
(I know that there are infinite domains with unity, like $\mathbb Z$ having finitely many subdomains with unity ($\mathbb Z$ has $0$ such subdomains) , so I relaxed the condition that the subdomains need not have unity.)
As Joel92 notes, for some $p$, $\mathbb{Z}_p \subseteq R$. Let $a\in R\setminus \mathbb{Z}_p$. If $a$ is transcendental over $\mathbb{Z}_p$, then you have infinitely many subdomains of the form $ \mathbb{Z}_p[a^n] $. Thus $a$ is integral over $ \mathbb{Z}_p$. It means that $R$ is a finite field extension of $\mathbb{Z}_p$, and is in particular finite.
In fact you can do better, since any finite extension that has finitely many subfields is a simple extension. Thus, $R$ is of the form $ \mathbb{Z}_p(\alpha) $, for some algebraic $\alpha$.