Closure of a Connected Set is connected proof The following is my attempt, but it seems far too simple to be correct. Could someone tell me where my logic is incorrect? 
Suppose $\bar{E} $ is not connected, and write $\bar{E} = E \cup \partial E$. 
If $\bar{E}$ were not connected, then for all $x \in E^{'}$ ,  $x \notin \partial E$, since we no limit point may be in $\partial E$.
But, if $x \in \partial E$, $x \in E^{'}$,  a contradiction. So, $\bar{E}$ must be connected. 
 A: You are right about the proof being pretty straightforward, however, I think your proof is not quite right. Assuming $\overline{E} = E \cup \partial E$ is too specific of a choice of sets to express $\overline{E}$. One definition of a connected set is a set that is not the union of two separated sets. So to assume $\overline{E}$ is not connected, you have to assume there are two sets $A,B$ such that $\overline{A} \cap B = A \cap \overline{B} = \emptyset$ and $\overline{E} = A \cup B$. Hence, you shouldn't just assume $\overline{E}$ looks like $E \cup \partial E$; you should show that no such $A,B$ exist, whether or not $A$ or $B$ is equal to $E$ or $\partial E$.
Here is how I would recommend proceeding:

By assumption $E$ is connected, so it cannot be expressed as the union of two separated sets. For the sake of contradiction, suppose $\overline{E}$ is not connected. Then we can find separated sets such that $\overline{E} = A \cup B$. However, you can show that $A\cap E$ and $B \cap E$ are disconnected sets. Lastly we observe that  $(A\cap E)\cup (B \cap E) = E$, implying $E$ is not connected.

Two other results that seem relevant to this proof are:

If $C,D$ are sets, $C$ is connected and $C \subset D \subset \overline{C}$ then $D$ is connected.

And

If $C$ is connected and $C \subset E \cup F$ where $E,F$ are separated, then either $C \subset E$ or $C \subset F$. 

If you can prove the former result, your proof follows as a special case of letting $D = \overline{E}$.
