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.
 A: I think it's simplest to see from the first definition.
Let $\mathcal{A}$ be the collection of closed sets containing $A$ and $\mathcal{B}$ the collection of closed sets containing $B$. Since $A \subset B$, we know $\mathcal{B} \subset \mathcal{A}$, and so $\bigcap \mathcal{A} \subset \bigcap \mathcal{B}$ (i.e. $\overline{A} \subset \overline{B}$).
Loosely speaking, adding more sets to an intersection can only make it smaller.
A: I think it's much simpler than that.  By definition #1, the closure of A is a subset of any closed set containing A; and the closure of B is certainly a closed set containing A (because it contains B, which contains A).  QED.
A: You say that some of the proofs you have looked use the argument "that since $A$ is contained in $\overline{B}$, then $\overline{A}\subseteq\overline{B}$" and that they don't seem strong enough for you but this follows directly from definition #1. Any closed subset containing $B$ contains $A$ and consequently $A\subseteq \overline{B}$. Since $\overline{B}$ is closed, $\overline{A}\subseteq\overline{B}.$
A: 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 $\bigcap \mathcal{C}_A \subseteq \bigcap \mathcal{C}_B$.
A: Using the first theorem from the question which is given as a definition in Rudin.
Theorem 1: $ \overline{A} = A \cup A' $, where $A'$ = the set of all limit points of $A$ (Definition 2.26 in Rudin). 
 
Since $A \subset B$, we have that $A \subset B \cup B'$ or $A \subset \overline{B}$

Next consider a point $x \in A'$ or $x$ is a limit point of $A$. This gives that every neighbourhood of $x$ contains a point $y \neq x$ such that $y \in A$ (Definition 2.18. (b) in Rudin). But this $y \in B$ since $A \subset B$ and hence we have that $x$ is a limit point of $B$ or $x \in B'$ or $x \in B\cup B'$ or $x \in \overline{B}$.  We have shown that  if $x \in A'$ then $x \in \overline{B}$ or $A' \subset \overline{B}$.
Since $A \subset \overline{B}$ and $A' \subset \overline{B}$ we have $A \cup A'=\overline{A} \subset \overline{B}$ 
QED.
A: I think there is one more important portion that is missing - we need to state (if not rigorously prove) that the closure $\overline{A} = A\cup L$ is in fact a closed set. Otherwise, as Stephen Abbott pointed out in his book Understanding Analysis, the union of sets could potentially create new limit points. That is, we need to prove that if $x$ is a limit point of $\overline{A}$, then $x$ is infact a limit point of $A$. To see this, we consider $\epsilon$-neighbourhood of $x$ and show that it contains some point in $A$ that is not equal to $x$. 
A: Consider a point $x \in A'$. This gives that every neighborhood of $x$ contains a point $y \neq x$ such that $y \in A$. But this $y \in B$ since $A \subset B$ and hence we also have that $x \in B'$ (because $x$ has neighborhood in $B$ that is not $x$).
Now we have that:
$A \subset B$ and $A' \subset B'$
$(A \cup A') \subset (B \cup B')$
$\overline{A} \subset \overline{B}$
