If $X \setminus A$ is disconnected then prove or disprove $X \setminus B$ is also disconnected Let $X$ be a connected metric space ( with more than one point ) and $A \subseteq X$ be not closed in $X$ and  such that $X \setminus A$ is not connected ; then is it true that $X \setminus B$ is also not connected for some closed ( in $X$ )  subset $B$ of $A$ ? 
 A: Yes, in metric spaces this is true. More generally, it holds in spaces satisfying the $T_5$ separation axiom - any to separated sets can be separated by open sets.
That $X\setminus A$ is not connected means there are two nonempty relatively open subsets $U_r,V_r$ of $X\setminus A$ such that $U_r\cap V_r = \varnothing$ and $X\setminus A = U_r\cup V_r$.
That $U_r$ and $V_r$ are both relatively open in $X\setminus A$ and disjoint means they are separated in $X$, i.e. $U_r \cap \overline{V_r} = \varnothing = \overline{U_r} \cap V_r$. By the $T_5$ axiom, there are disjoint open (in $X$ sets $U,V$ with $U_r\subset U$ and $V_r\subset V$. Then $B := X \setminus (U\cup V)$ is closed in $X$, and since $X\setminus A = U_r\cup V_r\subset U\cup V$, we have $B\subset A$. And by construction $X\setminus B = U\cup V$ is not connected.
We show that metric spaces are $T_5$-spaces: Let $E,F$ be disjoint sets in the metric space $X$, i.e. $E\cap \overline{F} = \varnothing = \overline{E}\cap F$. If either of the two is empty, a separation by open sets is trivial - we can choose one of the open sets as $\varnothing$. So suppose that both are nonempty. Then define
$$U := \{ x\in X : \operatorname{dist}(x,E) < \operatorname{dist}(x,F)\} \quad\text{and}\quad V := \{ x\in X : \operatorname{dist}(x,F) < \operatorname{dist}(x,E)\},$$
where $\operatorname{dist}(x,M) := \inf \{ d(x,y) : y \in M\}$. Since $E\cap \overline{F} = \varnothing$, for every $e\in E$ there is an $r > 0$ such that $B_r(x) \cap F = \varnothing$, so $\operatorname{dist}(e,E) = 0 < r \leqslant \operatorname{dist}(e,F)$, whence $E \subset U$. By the same reasoning $F\subset V$, and clearly $U\cap V = \varnothing$. It remains to see that $U$ and $V$ are open. By symmetry, it suffices to show that $U$ is open. Suppose $x\in U$. Then
$$s := \operatorname{dist}(x,E) < t := \operatorname{dist}(x,F)$$
by definition of $U$, and with $r := \frac{t-s}{3}$ we have $r > 0$ and for all $y\in B_r(x)$
$$\operatorname{dist}(y,E) \leqslant \operatorname{dist}(x,E) + d(x,y) < s + r < t - r < \operatorname{dist}(x,F) - d(x,y) \leqslant \operatorname{dist}(y,F),$$
so $B_r(x) \subset U$ and $U$ is recognised as open.
A: Is it true that $X \setminus B$ is also not connected for some closed ( in $X$
 ) subset $B$
 of $A$? 
The answer is: Generally it is not true.
Consider $X=R$ with topology $\mathcal{T}$  with base  $$\mathcal{B}=\left\{\left(-\infty,a\right) : \: a \in \mathbb{R}\right\}.$$
Let $a, x\in \mathbb{R}$ with $a<x$, consider  $A=\left[a,x\right) \cup \left(x, \infty\right)$, note that $A$ is not closed in $X$ and $X\setminus A= \left(-\infty,a \right) \cup {x}$ is not connected.
Note that if $B$ is closed (in $X$) subset of $A$, then there is  $y \in \mathbb{R}$ with $x<y$ such that $B=\left[y,\infty\right)$, furthermore, $X \setminus B$ is connected.
