If $V$ is open, then $V \cap \overline B \subseteq \overline{V \cap B}$

Question

I've read a proof of below lemma in this note.

Let $E$ be a topological space and $V,B \subseteq E$ such that $V$ is open. Then $V \cap \overline B \subseteq \overline{V \cap B}$.

The proof is

The inclusion $V \cap \overline{B} \subseteq \overline{V \cap B}$ used above holds in any topological space, where $V$ is open. For then the complement $W$ of $\overline{V \cap B}$ is open, and $W \cap V \cap B=\varnothing$. And since $W \cap V$ is open, this implies $W \cap V \cap \overline{B}=\varnothing$, which is another way to state the desired inclusion.

I tried to give it a more direct proof. Could you have a check on my attempt?

I posted my proof separately so that I can accept my own answer and thus remove my question from unanswered list. If other people post answers, I will happily accept theirs.

Answer 1

Fix $x\in V \cap \overline B$ and let $U$ be a neighborhood (nbh) of $x$. Because $V$ is open, $V\cap U$ is also a nbh of $x$. On the other hand, $x\in \overline B$, so $(V\cap U) \cap B \neq\emptyset$. This implies $U \cap (V\cap B) \neq \emptyset$ and thus $x \in \overline{V \cap B}$. This completes the proof.