general topology, subsets and limit points 
If $A, B$ are nonempty sets, and $A \subset B$, prove that $A' ⊂ B'$.

where $A'$ and $B'$ are the derived sets of $A$ and $B$ respectively
My attempt at the solution is this: let $x$ belong to $A$, then $x$ also belongs to $B$ since $A ⊂ B$. Then $x$ is a limit point to $A$ and thus belongs to $A'$ since $A'$ is the set of all limit points of $A$. And therefore $x$ is also a limit point to $B'$ so $B'$ and $A'$ have the same x and thus $B' \subset A'$.
Please if anyone could improve my proof or make a better one?
 A: The proof is not correct. What you should do is to take an element $x\in A'$ and then to prove that $x\in B'$ (which is easy).
A: Suppose $x\in A'$. Then, either $x\in A$ or $x$ is a limit point of $A$

If $x\in A\subset B\subset B'$, we are clearly done.
If $x$ is a limit point of $A$, there exists a sequence of points $a_1,a_2,...$ such that $\forall i\in\mathbb N,\; a_i\in A,\; d(a_i,x)<\frac1n$. But, $\forall i\in\mathbb N$, $a_i\in B$. Hence, $x$ is also a limit point of $B$, and we are done.
A: $x\in A'\iff \forall U\in \nu(x), U\cap A_x=\emptyset$ where $U$ is a neighborhood that contain $x$ and $A_x$ means $A\setminus \{x\}$.

Let $x\in A'$ and let $U\in \nu(x)$. Then, by definition, $U\cap A_x\neq\varnothing $. But $A_x\subset B_x$. Therefore $U\cap B_x\neq \varnothing $ and thus $x\in B'$.
A: Let $x\in A'$. 
Then for every open set $U$ that contains $x$ there is an element $y\in A\cap U$ with $x\neq y$. 
But $A\subset B$ implies that $A\cap U\subset B\cap U$ so that also $y\in B\cap U$.
Proved is now that $x\in B'$ and this is enough.

Above $\subset$ is used as a notation for "subset of".
