Set operation allowed? 


Suppose (WLOG) instead, $x \in E$, but $x \notin A$. So $x \in B$ by assumption that $E \subset A \cup B.$ This proves $E \subset B$. Similar argument shows $E \subset A$. But also we cannot have $x \in E$ and $x \notin A$ or $x \notin B$ because this violates $E \subset A \cup B.$
I didn't use the connectedness of $E$ at all…so I am worried.
As a side note, is the operation allowed for any set? this is for any set?

$X \subset C \iff X \cup D \subset C \cup D$
$X \subset C \iff X \cap D \subset C \cap D$

 A: You have not proved  that $E\subseteq B$, merely that some $x\in E$ is in one of $A,B$ (this is trivial, and not entirely true: it could be that $E=\emptyset$, which you have tacitly excluded).
Instead, you can argue by contraposition as follows: suppose $E$ has a point in each of $A,B$. Then try to show directly from the definition that it follows that it is not connected.
As for your side question, certainly not. Just consider $D=X$ for the first one and $D=\emptyset$ for the second.
A: Consider $E=E\cap(A\cup B)$, which follows from the hypothesis. Then
$$
E=(E\cap A)\cup(E\cap B)
$$
and $(E\cap A)\cap(E\cap B)=E\cap(A\cap B)=\emptyset$.
Thus we have written $E$ as the disjoint union of two open sets in $E$, precisely $E\cap A$ and $E\cap B$. What does connectedness of $E$ imply at this point?

About the other part of the question, you are surely allowed to say
$$
X\subset C \implies X\cup D\subset C\cup D
$$
but the converse implication does not hold: just consider two disjoint non empty sets $X$ and $C$ and take $D=X\cup C$. Then $X\cup D\subset C\cup D$ (they're equal, actually), but $X$ is not a subset of $C$.
For the intersection the situation is similar; the counterexample is obtained with $D=\emptyset$.
