Let $A \subset B \subset X$ and suppose $A$ is dense in $B$. Show that $A$ is dense in $cl_X(B)$. 
Let $A \subset B \subset X$ and suppose $A$ is dense in $B$. Show that $A$ is dense in $cl_X(B)$.

According to the properties of subspace topology I have that since $A \subset B \subset X$, then $$cl_B(A) = B \cap cl_X(A).$$
Now in order for $A$ to be dense in $cl_X(B)$ I need to show that $cl_B(A) = cl_X(B)$, but $cl_B(A) = B \cap cl_X(A) $ so in essence I need to show that $$B \cap cl_X(A) = cl_X(B)?$$
The closure has quite a bit of different definitions... It seems that this question could also be done by showing that $A \cap E \ne \emptyset$ for every open $E \in cl_X(B)$?
How can I figure out which is the right one to use? My instinct says that the first definition is the one to go with since the question was asked on a chapter regarding subspace topology.
Approaching with the former I have that
$”\subset”$ Since $A \subset B$, then $cl_X(A) \subset cl_X(B)$ and thus $B \cap cl_X(A) \subset B \cap cl_X(B)$, but $B \subset cl_X(B)$ so $B \cap cl_X(B) = cl_X(B)$ and so $B \cap cl_X(A) \subset cl_X(B)$.
$”\supset”$ Take $x \in cl_X(B)$, then according to the properties of the closure I have that for every open nbhd $U_x$ the intersection $U_X \cap B \ne \emptyset$. Now if only $U_x \subset cl_X(A)$, then I would have that $x \in cl_X(A) \cap B$ which would finish the proof(?), but I don’t think that $U_x \subset cl_X(A)...$
Edit: I seem to have messed up already on $”\subset”$... Since $B \subset cl_X(B) \implies B \cap cl_X(B) = B \ne cl_X(B).$
 A: Let $O’:=O \cap \text{cl}_X(B)$ be any non-empty open subset of $\text{cl}_X(B)$. We must show it intersects $A$. (So $O$ is open in $X$; we apply the definition of the subspace topology).
I’ll do the hands on argument: pick $x \in O \cap \text{cl}_X(B)$.
As $x$ is in the closure of $B$ (in $X$) and $O$ is an $X$-open neighbourhood of $x$, we have some $b \in B \cap O$. The latter set is $B$-open (and non-empty) and so intersects $A$, as $A$ is dense in $B$. In particular $O’$ intersects $A$. It follows that $A$ is dense in $\text{cl}_X(B)$.
A: Let $U$ be some non-empty, open subset of $\text{cl}_X(B)$. By definition $U= V\cap \text{cl}_X(B)$, for some $V$ open in $X$. We want to show that $U\cap A\neq \emptyset$. However, $V\cap B\neq \emptyset$. If that wasn't the case we would have
$$
B\subset X\setminus V,
$$
and since the right-hand side is closed, this would imply
$$
\text{cl}_X(B)\subset X\setminus V,
$$
i.e., $\text{cl}_X(B)\cap V= U= \emptyset$, contradiction. As a result, $U\cap B= V\cap B\neq \emptyset$. But, $V\cap B$ is open in $B$, and $A$ is dense in $B$, thus
$$
A\cap U= A\cap U\cap B= A\cap V\cap B\neq \emptyset,
$$
which shows that $A$ intersects any open subset of $\text{cl}_X(B)$, i.e., $A$ is dense in $\text{cl}_X(B)$.
A: Let $F$ be a closed subset of $cl_X(B)$ which contains $A$. There exists a closed subset $G$ of $X$ such that $F=G\cap cl_X(B)$. $G\cap B$ is a closed subset of $B$ which contains $A$, implies that $G\cap B=B$, $B\subset G$, and $cl_X(B)\subset G$, and $F=cl_X(B)$.
