If $A$ is a subset of $B$, then the closure of $A$ is contained in the closure of $B$. I'm trying to prove something here which isn't necessarily hard, but I believe it to be somewhat tricky.  I've looked online for the proofs, but some of them don't seem 'strong' enough for me or that convincing.  For example, they use the argument that since $A\subset \overline{B} $, then $ \overline{A} \subset \overline{B} $.  That, or they use slightly altered definitions.  These are the definitions that I'm using:
Definition #1: The closure of $A$ is defined as the intersection of all closed sets containing A.
Definition #2: We say that a point x is a limit point of $A$ if every neighborhood of $x$ intersects $A$ in some point other than $x$ itself.
Theorem 1: $ \overline{A} = A \cup A' $, where $A'$ = the set of all limit points of $A$.
Theorem 2: A point $x \in \overline{A} $ iff every neighborhood of $x$ intersects $A$.
Prove: If $ A \subset B,$ then $ \overline{A} \subset \overline{B} $
Proof: Let $ \overline{B} = \bigcap F $ where each $F$ is a closed set containing $B$.  By hypothesis, $ A \subset B $; hence, it follows that for each $F \in \overline{B} $, $ A \subset F \subset \overline{B} $.  Now that we have proven that $ A \subset \overline{B} $, we show $A'$ is also contained in $\overline{B} $.
Let $ x \in A' $.  By definition, every neighborhood of x intersects A at some point other than $x$ itself.  Since $ A \subset B $, every neighborhood of $x$ also intersects $B$ at some other point other than $x$ itself.  Then, $ x \in B \subset \overline{B} $.
Hence, $ A \cup A' \subset \overline{B}$.  But, $ A \cup A' = \overline{A}$.  Hence, $ \overline{A} \subset \overline{B}.$
Is this proof correct?
Be brutally honest, please.  Critique as much as possible.

Using Definition #1 makes it quite easy.  For each $A \subseteq X$, let $\mathcal{C}_A = \{ F \subseteq X : F\text{ is closed and }A \subseteq F \}$.  Then by Definition #1 it follows that $\overline{A} = \bigcap \mathcal{C}_A$.
Note that if $A \subseteq B$, then $\mathcal{C}_B \subseteq \mathcal{C}_A$, and therefore $\bigcap \mathcal{C}_A \subseteq \bigcap \mathcal{C}_B$.
