Suppose that $A$ and $B$ are subsets of $X$ such that $A \subseteq B$ then $Int(A) \subseteq Int(B)$. Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ and $B$ are subsets of $X$ such that $A \subseteq B$ then $Int(A) \subseteq Int(B)$.
I know this a true statement so now I need to go about proving it. I am in an introduction to proofs class and we are learning set theory and topology as a basis for prove writing.  I only have a basic knowledge of topology and typically our proofs involve the definition and sets/elements. 
My definition for interior is: Let $(X, \mathfrak T)$ be topological space and let $A \subset X$.  The interior of $A$ is the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U \subseteq A$.
Based on the definition I know that $Int (A) \subseteq A$.
Here is my proof:
Let $x \in Int(A)$ then $x \in (A)$ since $A \subseteq B$ then $x \in B$.. I feel like I am on the right track however I know that just because an element is in A does not necessary mean that it is in the interior of A since A is an open set.   

Your proof only shows interior point of A is in $B$, but doesn't show it's interior point of $B$.
Let $x\in int(A)$. Then by definition, there exists an open set $U$ such that $x \in U \subseteq A$.Sinc $A\subset B$, hence $x \in U \subseteq B$, then by definition, $x\in int(B)$.
