Proving disconnected sets. I came up with two weak incomplete solutions to the problem.

If $E$ and $F$ are connected subsets of a metric space $M$ with $E \cap F \neq \emptyset$. Show $E \cup F$ is connected.

failed proof1:
If $E \cup F$ is not connected, let $W$ and $V$ be disjoint nonempty open sets that separate $E \cup F = W \cup V$. So if $x \in E \cap F \subset E\cup F = W \cup V$. First neither $W, V$ can equal either $E$ or $F$, if it does we have the first contradiction. Notice the situation $W \subset E$ would mean that $E \cup F = V$, so the sets $W$ and $V$ would not be empty. So we are left with (without any loss of generality), $E \subset W$, then $E \not\subset V.$
if $x \in W, x \notin V$. The sets $V$ and $W$ are open relative to $E \cup F$. So we must have, 
$$W = (E \cup F) \cap O_m^1,$$
$$V = (E \cup F) \cap O_m^2$$
for open sets $O_m^1, O_m^2 \in M$.
So $\emptyset = E \cap V = E \cap [(E \cup F) \cap O_m^2 ] = E \cap F \cap O_m^2 \neq \emptyset$, contradiction.
faled proof2:
Construct a continuous function (this is automatically surjective) $f: M \to M$. Assuming $E\cup F$ is disconnected, the image $f(E \cup F) = f(E) \cup f(F)$. I got nowhere with this.
 A: I think you are closer with your first attempt. In my experience, proofs regarding connectedness are often done with contradiction. I agree with where you are going with the proof up to where you state "$E \cup F = W \cup V$".After this point, I feel that you introduce some unnecessary cases. One nice result to apply here instead is that: 
If $C$ is a connected subset and $C \subset A \cup B$ where $A \cap B = \emptyset$ then either $C \subset A$ or $C\subset B$.
Applying that to what you have, it is obvious that $E \subset W \cup V$ so (without loss of generality) we know $E \subset W$ and $E \cap V = \emptyset$. Similarly $F \subset W \cup V$. $F$ is also connected, so $F \subset W$ or $F \subset V$. We can now introduce cases.
Case 1: $F \subset W$ and $F \cap V = \emptyset$
Then $E\cup F = W$ so $E\cup F = W \cup V$ implies that $V = \emptyset$ which is a contradiction as you already noted. 
Case 2: $F \subset V$ and $F \cap W = \emptyset$
Then $$(E \cap F) \subset (W \cap V) = \emptyset$$ so $E \cap F = \emptyset$ which is also a contradiction. We now conclude that $E \cup F$ is connected.
A: If $E\cup F=V\cup W$ for $V,W$ nonempty, open and disjoint, then $V\cap E$ and $W\cap E$ splits $E$ into two open disjoint parts, so either $V\supseteq E$ or $V\subseteq F\setminus E$. Similarly with $V,F$ resp. $W,E$ and $W,F$. All these cases lead to contradictions: if, for example, $V\supseteq E$, then $W\subseteq F\setminus E$, so it is not true that $W\supseteq F$, so it must be $W\subseteq E\setminus F$, hence $W=\emptyset$.
