Let $B$ be a subset of a space $X$. Prove that the following statements are equivalent 
Question: Let $B$ be a subset of a space $X$. Prove that the following statements are equivlent.$^1$
(a) $B$ is nowhere dense.
(b) $X\setminus \bar{B}$ is dense in $X$.
(c) $X\setminus \overline{(X\setminus \bar{B})}=\emptyset$.
(d) $B\subseteq \overline{(X\setminus \bar{B})}$.

Notation:

*

*$\operatorname{int}A$ is an interior of $A$.

*$A'$ is a derived set of $A$.

My attempt

*

*(a)$\Rightarrow$(b): If $B$ is nowhere dense, then $\operatorname{int}\bar{B}=\operatorname{int}(X\setminus (X\setminus \bar{B}))=\emptyset$. Thus $X\setminus \bar{B}$ is dense in $X$.

*(b)$\Rightarrow$(c): If $X\setminus \bar{B}$ is dense in $X$, then $\overline{(X\setminus \bar{B})}=X$, so $X\setminus \overline{(X\setminus \bar{B})}=X\setminus X=\emptyset$.

*(c)$\Rightarrow$(d): If $X\setminus \overline{(X\setminus \bar{B})}=\emptyset$, then $\overline{(X\setminus \bar{B})}=X$, so $B\subseteq X = \overline{(X\setminus \bar{B})}$.

*(d)$\Rightarrow$(a): Suppose that $B\subseteq \overline{(X\setminus \bar{B})}$. Then $\bar{B}\subseteq\overline{(X\setminus \bar{B})}$ and so $\operatorname{int}\bar{B}\subseteq\operatorname{int}\overline{(X\setminus \bar{B})}$. Suppose that $\operatorname{int}\bar{B}\ne\emptyset$. Then there exists $x\in \operatorname{int}\bar{B}$, and so $x\in\operatorname{int}\overline{(X\setminus \bar{B})}$. Since $x\in\operatorname{int}\overline{(X\setminus \bar{B})}$, $x\in (X\setminus \bar{B})$ or $x\in(X\setminus \bar{B})'$. If $x\in (X\setminus \bar{B})$, then $x\in\operatorname{int}\bar{B}\subseteq\bar{B}$ and $x\in (X\setminus \bar{B})$, which is a contradiction. If $x\in(X\setminus \bar{B})'$, then $\operatorname{int} \bar{B}$ must contain an element of $X\setminus \bar{B}$, but $\operatorname{int}\bar{B}\cap (X\setminus \bar{B})=\emptyset$, a contradiction. Thus, $\operatorname{int}\bar{B}=\emptyset$ and so $B$ is nowhere dense.

In my attempt, the proof that (d) implies (a) is much more longer than other three, so I feel strange. Are there any proofs to justify (d)$\Rightarrow$(a) shorter and easier than mine?

$^1$ Fred H. Croom(2003). Principles of Topology. Cengage Learning. page 108.
 A: For $(d)\Rightarrow(a)$, $\bar{B}\subseteq\overline{X\setminus\bar{B}}=X\setminus\text{int}\bar{B}=(X\setminus\bar{B})\cup (\bar{B}\setminus\text{int}\bar{B})\Rightarrow \bar{B}\subseteq\bar{B}\setminus\text{int}\bar{B}$, which means $\text{int}\bar{B}=\emptyset$.  (Note that the trivial $\overline{X\setminus A}=X\setminus\text{int}A$ is invoked.)
A: You can do it without fancy formulas.
(a)$\implies$(b) Let $U$ be an open set such that $U\cap(X\setminus \overline{B})=\emptyset$. Then $U\subseteq\overline{B}$; since $\overline{B}$ has empty interior, this implies $U=\emptyset$. So every nonempty open set intersects $X\setminus\overline{B}$.
(b)$\implies$(c) Since $\overline{X\setminus\overline{B}}=X$, the statement is obvious.
(c)$\implies$(d) Since $\overline{X\setminus\overline{B}}=X$, the statement is obvious.
(d)$\implies$(a) Let $U$ be a nonempty open set contained in $\overline{B}$. Then $U\cap B\ne\emptyset$, so we can take $x\in U\cap B$. By assumption, $x\in\overline{X\setminus\overline{B}}$, so there is $y\in X\setminus\overline{B}$ with $y\in U$, because $U$ is a neighborhood of $x$. Therefore $y\notin\overline{B}$, a contradiction.
