prove that if $E$ is connected and $E \subseteq F \subseteq \overline{E}$, then $F$ is connected. 
Define a set $A$ to be disconnected iff there exist nonempty relatively open sets $U$ and $W$ in $A$ with $U\cap W = \emptyset$ and $A = U\cup W.$ Define a set $A$ to be connected iff it is not disconnected.(there are many equivalent definitions, but I want to prove this lemma using this one). Prove that if $E$ is connected and $E\subseteq F \subseteq \overline{E},$ then $F$ is connected.

Let $U, W$ be a separation for $F$. Find open sets $O_U$ and $O_W$ so that $U = F \cap O_U$ and $W = F\cap O_W.$ I claim that $E\cap O_U, E\cap O_W$ separate $E$. However, I'm unable to show that $U' = E\cap O_U, W' =E\cap O_W \neq \emptyset$ (I think this should be straightforward, but for some reason, I can't figure this out). Suppose $U' = \emptyset.$ Then $E\cap O_U = \emptyset.$ Since $E \subseteq F = U\cup W = F\cap (O_U \cup O_W)\subseteq O_U\cup O_W,$ we have that $E \subseteq O_W,$ so $E\cap O_W = E\subseteq F\cap O_W = W\subseteq F\subseteq \overline{E}.$ Observe that since $E\cap O_U = \emptyset, F\cap O_U = (F\backslash E)\cap O_U\subseteq F\backslash E.$
Similarly, $W'\neq \emptyset.$ Clearly, $U', W'$ are relatively open in $E$. Suppose $U'\cap W' \neq \emptyset.$ Let $x\in U'\cap W'.$ Then $x\in E\cap O_U\cap O_W \subseteq F\cap O_U\cap O_W = U\cap W = \emptyset,$ a contradiction. So $U'\cap W' = \emptyset.$ Also, $U'\cup W' = (E\cap O_U)\cup (E\cap O_W) = E\cap (O_U \cup O_W)$ and $E\subseteq (O_U \cup O_W),$ so $U'\cup W' = E.$

Assume $E \subseteq F \subseteq \overline{E} \subseteq X$ and that we're working in the topology of $E$ relative to $X$.  If $F$ is disconnnected, then $\exists U, V \subseteq X$ open (in $X$) such that $F \subseteq U \cup V$ and $U \cap V \cap F = \emptyset$.
But then $U \cap V \cap E = \emptyset$ and $E \subseteq F \subseteq U \cup V$, so $U$ and $V$ are a separation for $E$, which is connected.  Thus, without loss of generality, $E \subseteq U$ and $E \cap V = \emptyset$.
But then $E \subseteq X \setminus V$, which is closed (because $V$ is open), and since the closure of $E$ is the intersection of all closed sets containing $E$, that means $\overline{E} \subseteq X \setminus V$, so $\overline{E} \cap V = \emptyset$ and since $F \subseteq \overline{E}$, we also have $F \cap V = \emptyset$ so that $U, V$ do not separate $F$.
