Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ and $B$ are subsets of X. If $A\subseteq B$ then $A' \subseteq B'$ Let $(X, \mathfrak T)$ be a topological space and suppose that $A$ and $B$ are subsets of X.
If $A\subseteq B$ then $Bd(A) \subseteq Bd(B)$
If $A\subseteq B$ then $A' \subseteq B'$  ($A'$ is the set of limit points)
Let $( X, \mathfrak T_U)$ be the topological space. Let $A = [0,1) \cup (1,2)$ Let $B = [0,1) \cup (1,3)$
Then $A\subseteq B$ but $Bd(A)= \{0,1,2\}$ and $Bd(B) = \{0,1,3\}$ therefore $Bd(A) \not \subseteq Bd(B)$ so this is a false conjecture. 
I am a little more uncomfortable with the limit points. I believe $A'= [0,1] \cup [2, \infty)$ and $B'= [0,1] \cup [3, \infty)$ and therefore again this would be a false conjecture.  
My definition of limit point is that every open set containing $x$ contains a point of $A$ different from $x$.  
My definition of boundary is: Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is in the boundary of $A $if every open set containing $x$ intersects both $A$ and $X−A$
 A: Remember the definition for $D'$, $D\subseteq X$:
$$D'=\{x\in X/\ \forall V\in \mathfrak T(x\in V\implies \exists d\in D(d\neq x \wedge d\in V))\}$$
(I'm sorry for the symbolic notation), let's prove that $A'\subseteq B'$ if $A\subseteq B$.
Let $x\in A'$ and take any $V\in \mathfrak T$ such that $x\in V$, since $x\in A'$ there exists $a\in A$ such that $a\neq x$ and $a\in V$, in particular, if we let $b:=a$, we see that $b\in B$ (since $A\subseteq B$), $b\neq x$ and $b\in V$. This shows that 
$$\forall V\in \mathfrak T(x\in V\implies \exists b\in D(b\neq x \wedge b\in V))$$
so $x\in  B'$. Since $x\in A'$ is arbitrary, we have that $A'\subseteq B'$.
A: If $a_n \to x$, $a_n \in A$ and $a_n \neq x$, then we also have $a_n \in B$.
A: Let $x$ be a limit point of $A$. Then for any open set $U$ with $x \in U$ we know $$\left[U\setminus \{x\}\right]\cap A \neq \emptyset$$ Since $A\subset B$ then $$\left[U\setminus \{x\}\right]\cap A \subset \left[U\setminus \{x\}\right]\cap B$$ and hence $$\left[U\setminus \{x\}\right]\cap B \neq \emptyset$$ so every limit point of $A$ is a limit point of $B$.
