So I am aware that when a subset $B$ is dense in another subset $U,$ we could mean either $U \subset \overline{B}$ or $B \cap U$ is of some subspace $U,$ a dense subset.
But what I was wondering is, how can we formally show that if $B$ is dense subset of $X$ and $U$ is given to be open, then $B\cap U$ is also dense in $U?$
I would greatly appreciate the help, as I think this will help me conceptually better understand this section on subspaces and dense subsets.
On
Suppose $V$ is open in $X$ and $V\subseteq U$. Then, $V$ contains a point of $B$ which is necessarily in $B\cap V$ and $B\cap V\subseteq B\cap U$. Thus every open subset of $U$ contains a point of $B\cap U$.
On
A general fact: if $U$ is a subspace of $X$ we have two notions of closure, in $X$ and in $U$ (in the subspace topology) which are nicely related: if $B \subseteq U$ both closures would make sense and in fact
$$\operatorname{Cl}_U(B) = \operatorname{Cl}_X(B) \cap U$$
Now your question is answered:
Such a $B$ is dense in $U$ iff $\operatorname{Cl}_U(B) = U$ iff $\operatorname{Cl}_X(B) \cap U = U$ iff $U \subseteq \operatorname{Cl}_X(B)$.
So you want to show that $U \subset (B \cap U)^{-}$. Let $x \in U$. To show that $x$ is in the closure of $B \cap U$ you have to show that if $V$ is any open set containing x then $V$ intersects $B \cap U$. But $U \cap V$ is open and it conatins $x$ so it is a non-empty open set. Since $B$ is dense it intersects every non-empty open set, so $B \cap U \cap V$ is non-empty which is what we wanted to prove.