Connecting two definitions of connectedness Let $E$ be a subset of a metric space $X$, and say $E$ is connected if and only if it is not disconnected.  I have the following two definitions of disconnected:
I)  $E$ is disconnected if there exist two nonempty subsets $A$, $B$ of $X$ such that:


*

*$E = A \cup B$

*$A \cap \overline{B} = \emptyset$ and $\overline{A} \cap B = \emptyset$


II) $E$ is disconnected if there exist two open (relative to the metric space $X$) subsets $A$, $B$ of $X$ such that:


*

*$E \subset A \cup B$

*$E \cap A \neq \emptyset$ and $E \cap B \neq \emptyset$

*$E \cap A \cap B = \emptyset$


Is there a concise way to prove that these definitions are equivalent?
Here are the definitions I'm using:
A neighborhood of radius $r > 0$ around a point $p \in X$ is the set $N_r(p) = \{q \in X \; | \; d(p, q) < r\}$.  A limit point of a subset $E$ of $X$ is a point $p$ such that every neighborhood of $p$ contains at least one other point of $E$.  A subset $E$ of $X$ is closed if it contains all of its limit points.  The closure of $E$, written $\overline{E}$, is the set $E$ together with its limit points.  A subset $E$ of $X$ is open if, for every point $p \in E$, there is some neighborhood of $p$ which is contained in $E$.
Assume we have already proved that every closure is closed, and that a set is closed if and only if its complement is open.
 A: I$\implies$ II: Suppose we have such sets $A,B$. The sets we use are $E\setminus \bar{A}$ and $E\setminus \bar{B}$, which are open. Then $A\subseteq E\setminus \bar{B}$ and $B\subseteq E\setminus \bar{A}$, so $E\setminus \bar{A}\cup E\setminus \bar{B}=E$ thus 1 is satisfied. Neither $E\cap A$ nor $E\cap B$ are empty since $A,B\subset E$, so $E\setminus \bar{B}$ and $E\setminus \bar{A}$ are nonempty thus 2 is satisfied. Furthermore, $$E\cap(E\setminus \bar{A})\cap (E\setminus \bar{B})=E\setminus(\bar A\cup\bar B) = E\setminus E=\emptyset$$ thus 3 is satisfied.
II$\implies$I: Suppose we have such sets $A,B$. The sets we use are $E\cap A$ and $E\cap B$. Then $E=(A\cap E)\cup (B\cap E)$ so 1 is satisfied, and $(E\cap A)\subseteq E\setminus E\cap B$ so since $E\cap A$ is open in $E$ we have $(E\cap A)\subseteq E\setminus \overline{E\cap B}$ thus $(E\cap A)\cap \overline{E\cap B}=\emptyset$, and the other intersection is similar, so 2 is satisfied.
A: Assume firstly: If $E$ is disconnected as Definition-I sense, 
There exists $A$ and $B$ some subsets of $X$, such that  $$E= A\cup B\text{ and } A\cap \bar{B}=\phi,\bar{A}\cap B=\phi$$
These condition implies that $A$ and $B$ are open subset in $X$, as $A$ and $B$ are subset of $X$ and $\bar{A},\bar{B}$ are closed sets and complement of closed set is open.
Hence we have two non empty open set $A$ and $B$ which will satisfy condition-II (verify it).  Hence $E$ is disconnected in sense of definition-II.
Conversely,  if we have given condition of definition-II:  
That is we have given $A$ and $B$ open set satisfying the definition-II, Take $A'= E\cap A$ and $B'= E\cap B$, $A'$ and $B'$ will satisfy the condition of definition-I (verify it).
Hence $E$ is disconnected in sense of definition-1.
