Let $R$ be an integral domain such that every proper subring (with unity) of it is a PID ; then is $R$ a PID? 
Let $R$ be an integral domain such that every proper subring (with unity) of it is a PID ; then is it true that $R$ is a PID ?

Relevant definitions
Integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero.
PID is an integral domain in which every ideal is principal, i.e., can be generated by a single element.
 A: Yes, essentially because there are virtually no such rings $R$ at all.  Indeed, a domain $R$ has the property that every proper subring is a PID iff either $R$ is an algebraic extension field of $\mathbb{F}_p$ for some $p$ or $R$ is a subring of $\mathbb{Q}$.  (In fact, the proof will only use the assumption that every proper subring of $R$ is integrally closed.)
First, suppose $R$ has positive characteristic $p$.  If there is some element $x\in R$ that is transcendental over $\mathbb{F}_p$, then the subring $\mathbb{F}_p[x^2,x^3]\subset R$ is not a PID.  So every element of $R$ is algebraic over $\mathbb{F}_p$.  It follows that $R$ is a field, and in fact an algebraic extension of $\mathbb{F}_p$.
Now suppose $R$ has characteristic $0$.  Again, $R$ can have no transcendental elements: if $x\in R$ is transcendental over $\mathbb{Z}$, the subring $\mathbb{Z}[x^2,x^3]\subset R$ is not a PID.  Now suppose $a\in R$ and $a\not\in\mathbb{Q}$.  Let $f(t)\in\mathbb{Z}[t]$ be the polynomial obtained by taking the minimal polynomial of $a$ over $\mathbb{Q}$ and multiplying by the least common multiple of the denominators of its coefficients.  Note that $f(t)$ generates the ideal of polynomials in $\mathbb{Z}[t]$ which vanish at $a$ (this follows from Gauss's lemma) and $\deg f>1$ (since $a\not\in\mathbb{Q}$).  Let $p\in\mathbb{Z}$ be a prime that does not divide the leading coefficient of $f(t)$.  I claim that the subring $\mathbb{Z}[pa]\subseteq R$ is proper and is not a PID.  Indeed, it suffices to show $a\not\in \mathbb{Z}[pa]$, since then $\mathbb{Z}[pa]$ is not integrally closed.
If $a\in \mathbb{Z}[pa]$, then there is some polynomial $g(t)\in\mathbb{Z}[t]$ such that $g(pa)=a$, so $f(t)\mid g(pt)-t$ in $\mathbb{Z}[t]$.  Taking this mod $p$, we find that $f(t)\mid n-t$ in $\mathbb{F}_p[t]$, where $n$ is the constant term of $g(t)$.  But this is impossible, since the leading coefficient of $f$ is not divisible by $p$ and $\deg f>1$.
Thus $R$ must be a subring of $\mathbb{Q}$.  Any such subring is a localization of $\mathbb{Z}$ and in particular is a PID.
