Adding to the Answer provided by Hemant,
Option C : False
"If all finite subsets of $L$ are regular, then $L$ is regular."
Every finite language is (trivially) Regular. So, Every Finite subset of a language $L$ indeed always is Regular. So, This Hypothesis that "If all finite subsets of $L$ are regular" is Always True regardless of what $L$ is. $L $ could be any language, It could also be Non-regular.
Option B : True
"If all proper subsets of $L$ are regular, then $L$ is regular."
This is True and we will see "Why", by taking a more general result which implies that "If all proper subsets of $L$ are regular then $L$ is Finite and Hence Regular."
Claim : "every infinite language $L$ has a non-regular subset"
Proof :
Case 1 : If $L$ is Non-regular :
Then Since $L$ itself is a subset of $L$, we can say that the claim holds good. Moreover if we remove Finite number of strings from $L$, It will still remain Non-regular.
Case 2 : If $L$ is Regular :
Then $L$ will satisfy Pumping lemma for Regular languages which states that
If $L$ is Regular then $\exists P \geq 1$, such that $\forall $ strings $w \in L,$ where $|w| \geq P$, $\exists x,y,z,$ such that $w = xyz$ and $|xy| \leq P$ and $|y| \geq 1$ and $\forall q \geq 0$, $xy^qz \in L$
So, Since $L$ is Regular and Infinite, let's pick any random string $w = xyz$ in $L$ such that $|w| \geq P$, then by pumping lemma, all strings of the form $xy^qz$ will be in $L$. So, Let's take a subset $S$ of $L$ which is
$S = \left \{ xy^qz| q\,\, is\,\, Prime \right \}$
Clearly this subset $S$ of $L$ is Not Regular.
Hence, Our Claim that "Every Infinite language has a Non-regular subset" is True.
The above result implies the following :
If all proper subsets of $L$ are regular, then $L$ is Finite and hence, Regular.
