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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 ?

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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.

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    Haha, makes sense. Beautiful answer! When I read the question, this approach using algebraic elements didn't occur to me at all, but reading your answer it seems very natural. – 2017-02-05
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    @Eric Wofsey : how does $f(t)$ generates the ideal of polynomials of $\mathbb Z[t]$ that vanishes at $a$ (I am even unable to see that the ideal of polynomials vanishing at $a$ is principal ) ? – 2017-02-05
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    @SaunDev: First, $f(t)$ generates the ideal of polynomials in $\mathbb{Q}[t]$ which vanish at $a$, since it is a constant multiple of the minimal polynomial. Now suppose $g(t)\in\mathbb{Z}[t]$ vanishes at $a$. Then $f(t)$ divides $g(t)$ as elements of $\mathbb{Q}[t]$, so $g(t)=f(t)h(t)$ for some $h(t)\in\mathbb{Q}[t]$. Since $f(t)$ has trivial content, this implies $h(t)$ is actually in $\mathbb{Z}[t]$ by Gauss's lemma. – 2017-02-05
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    @EricWofsey : what happens if we ask for those rings $R$ such that every proper subring with unity is a PIR (principal ideal ring ) ? – 2017-08-06
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    @misao: Then the answer is no: consider $\mathbb{F}_p[x,y]/(x^2,y^2,xy)$. Completely classifying such rings seems substantially harder than the case of PIDs. – 2017-08-06