Several simple propositions about regular languages
In all following counterexamples, $A$ and $B$ are subset of the full language $F=\{0,1\}^*$. Note that $F$ is a regular language.


  If $A$ is regular pumpable $B\subseteq A$, then $B$ is regular pumpable too.


Counterexample: $A=F$ while $B$ is any non-regular-pumpable language. 

(It is taken for granted that a language being regular-pumpable means it is able to be pumped like a regular language. There are different ways to define being able to be pumped like a regular language; however it is defined, $\{0^n1^n\mid n\ge 0\}$ will not be regular pumpable. I made this note since some non-regular languages can be pumped like a regular language. For example, $\{ 01^{n^2} \mid n \ge 1 \} \cup \{ 0^kw \mid k\neq 1, w\in \{1\}^* \}$ is non-regular but can be pumped by the most common version of the pumping lemma for regular languages.)

(In case "regular pumpable" for a language is defined simply as "regular", just let $A=F$ and B be any non-regular language for a counterexample.) 


  If $A$ not regular pumpable and $A\subseteq B$, then $B$ is not regular pumpable too.


Counterexample: $B=F$.


  If $A$ is regular and $B\subseteq A$, then $B$ is regular too.


Counterexample: $A=F$ while $B$ is non-regular.


  If $A\cap B$ is regular, then $A$ and $B$ are regular too.


Counterexample: $A$ is not regular while $B=F\setminus A$. $A\cap B=\emptyset$ is regular.


  There is at least one finite language on the alphabet $\{0,1\}$, that doesn't meet the pumping lemma rules.


That is not true. All finite languages are regular and, thus, meeting the pumping lemma rules.




  How do you come to that conclusion?


Mostly experience and knowledge plus a bit of luck. Before proving any proposition, you would like to do some sanity check with some simple and/or extreme cases. For example, you might have noticed that I have used the simple and extreme language $F$ several times.