If $E$ and $F$ are connected subsets of $M$ with $E\cap F\ne\emptyset$, show that $E\cup F$ is connected. If $E$ and $F$ are connected subsets of $M$ with $E\cap F\ne\emptyset$, show that $E\cup F$ is connected.
My attempt:
Suppose $E\cup F$ is disconnected. $\boldsymbol{\Rightarrow \exists} \ open \ sets\ A, B\ne \emptyset$ such that $A\cap B=\emptyset$ and $A\cup B=E\cup F.$
Consider $E_1=A\cap E$ and $E_2=B\cap E$.  
$A\cap B=\emptyset\ \boldsymbol{\Rightarrow} (A\cap E)\cap (B\cap E)=\emptyset.$ 
$A\cup B=E\cup F\ \boldsymbol{\Rightarrow}\ (A\cap E)\cup (B\cap E)=E$
claim: $A\cap E\ne \emptyset$ and $B\cap E\ne \emptyset$.
Suppose $A\cap E=\emptyset \boldsymbol{\Rightarrow} B=E \boldsymbol{\Rightarrow}A=F$. This is a contraction since $E\cap F\ne \emptyset$.
Maybe I should prove $E_1 $ and $E_2$ are open in E. I don't know how to do next.

Your proof is not valid, for you need nonempty open sets that disjoint and cover $E$ to show that $E$ is disconnected. You've produced disjoint sets, but either might be empty, and in the case that $E$ and $F$ are the upper and lower semicircles of the unit circle within the plane, the intersection of $A$ with $E$ will be a subset of the upper semicircle, which will never be an open set in the plane unless it is the empty set. 
