If A⊂B,prove that $A^\circ⊂B^\circ$ and $cl (A) ⊂ cl(B)$ For the open sets,  $A ⊂ B,  B^\circ ⊂ B,  A^\circ ⊂ A$  We can say $A^\circ ⊂ B$ ..not sure what to do next.
For the closed sets, $A ⊂ cl(A)  , B ⊂ cl (B)$ then $A ⊂ cl (B)$ and I'm not sure what do next?
Am I on the right track or I need to use the definitions for the proof?
 A: Well ...

... or [do] I need to use the definitions for the proof?

yes. There might be some shortcut (which then would use the definition), but even then it's a good practice to use the definitions.
So, if you don't want to miss the practice, don't read the following spoiler.
In the first exercise it's near an immediate deduction from the very definitions:
\begin{align*}
   A^° = \bigcup_{U\in\mathcal{O}(X), U\subseteq A} U
\end{align*}
therefore:
\begin{align*}
   x \in A^° \Rightarrow \exists U \in \mathcal{O}(X): U\subseteq A\wedge x \in U
\end{align*}
and since $A \subseteq B$ it holds that $\exists U \in \mathcal{O}(X): U\subseteq A\subseteq B\wedge x \in U$, so $x \in B^°$.
Even in the second exercise it's just a small step -- but you may want to go by contradiction to keep it simple:
\begin{align*}
   x \notin \bigcap_{C\in\mathcal{C}(X), B \subseteq C} \Rightarrow \exists C \in \mathcal{C}(X) : B \subseteq C \wedge x \in C
\end{align*}
and since $A\subseteq B$, also $A \subseteq C$ and therefore $x \notin cl(A)$.
A: Just use the definitions. I assume you are talking about subsets of $\mathbb{R}^n$.
Let us prove $A^\circ \subset B^\circ$. Let $x\in A^\circ$. Then there is $\epsilon>0$ such that $B(x,\epsilon) \subset A\subset B$. Therefore $x\in B^\circ$.
Now the other part. First $A\subset B\subset \mathrm{cl}(B)$. The closure of $A$ is the intersection of all closed sets containing $A$. Since $\mathrm{cl}(B)$ is a closed set containing $A$, we get $\mathrm{cl}(A)\subset \mathrm{cl}(B)$.
A: $ x \in A^\circ $ <=> there exists an open set $O_x$ containing $x$ in $A$. So we have $O_x \subset A \subset B$. Hence, by definition, $x \in B^{\circ}$. 
For the other part we follows from the fact that : ${B^C}^\circ \subset {A^C}^\circ$.
Complement of these sets are closures of $B$ and $A$.   
A: Hint:
You can use the following equivalent definitions:
$A^o$ is the largest open set contained within $A$ and $cl(A)$ is the smallest closed set containing $A$. Have you proved these already / seen these yet?
