# Let $(X,\mathfrak T)$ be a topological space and suppose that A and B are subsets of X such that $A \subseteq B$ then $Cl(A) \subseteq Cl(B)$.

Let $(X,\mathfrak T)$ be a topological space and suppose that A and B are subsets of X such that $A \subseteq B$ then $Cl(A) \subseteq Cl(B)$.

My definition of closure is "Let $(X, \mathfrak T)$ be a topological space and let $A \subseteq X$. The closure of $A$ is defined by $Cl(A) =\bigcap \{U \subseteq X : U$ is a closed set and $A \subseteq U\}$. From the definition I know that $A \subseteq Cl(A)$.

Here is my proof: Let $x \in Cl(A)$ by the definition of closure $A \subseteq Cl(A)$ therefore $x \in A$. Since $A \subseteq B$. $x \in B$ then $x \in Cl(B)$ since $B \subseteq Cl(B)$ therefore $Cl(A) \subseteq Cl(B)$.

How does my proof look? All of the proofs on topologies I have been doing involve set theory which is why I started with an element of one set and tried to show it was an element in the other set.

This is not correct. The very first thing you say is that $x\in Cl(A)$, $A\subseteq Cl(A)$ implies $x\in A$... and it need not be. The closure of $A$ is contains $A$ itself, but may contain points not in the original set $A$.
Here, you need to use the definition of the closure: namely, that if $A\subseteq U$ and $U$ is closed, then $Cl(A)\subseteq U$ as well. (You can, and should, think of the closure as the smallest closed set that contains $A$.)
Now, consider $Cl(B)$. This is a closed set (it is the intersection of a collection of closed sets). Further, we know that $A\subseteq B$ and $B\subseteq Cl(B)$, so that $A\subseteq Cl(B)$.
So, $Cl(B)$ is a closed set that contains $A$, and therefore $Cl(A)\subseteq Cl(B)$, as desired.