Prove that if for $E$ and $F$ connected sets, if $E \cap F \neq \emptyset \rightarrow E \cup F$ is connected Prove that if $E \cap F \neq \emptyset \rightarrow E \cup F$ is connected.
I am trying to do it by contradiction.
Assume that $E \cup F$ is disconnected. Therefore, $\exists U, V$ open, such that
$$[(E \cup F) \cap U] \cap [(E \cup F) \cap V] = \emptyset$$
$$[(E \cup F) \cap U] \cup 
[(E \cup F) \cap V] = E\cup F$$
$$(E \cup F) \cap U \neq \emptyset$$
$$(E \cup F) \cap V \neq \emptyset$$
I am stuck with how to proceed.

Clearly $$(E \cap U) \cap (E \cap V) \subseteq ((E \cup F) \cap U)\cap ((E \cup F) \cap V) = \emptyset\tag 1$$ as $E \subseteq E \cup F$. The same holds for $F$.
Also $$(E \cap U) \cup (E \cap V) = E\tag 2$$ (left to right inclusion is clear, and if $x \in E$ it is in $E \cup F$ so in either $(E \cup F) \cap U$ (and thus in $E \cap U$) or in $(E \cup F) \cap V$ (and thus in $E \cap V$) for the other).
The same holds for $F$ instead of $E$.
Now, as $E$ is connected, we cannot have that both $E \cap U \neq \emptyset$ and $E \cap V \neq \emptyset$, (or else we'd have disconnection of $E$) so say for definiteness that we have $U \cap E = \emptyset$. This then implies that $E \cap V = E$ by (2), so that $E \subseteq V$. 
If now $p \in E \cap F$ (we have to use that too, of course) we see that $F \cap V \neq \emptyset$, and again connectedness of $F$ implies that $F \cap U = \emptyset$ (or $F \cap U$ and $F \cap V$ would otherwise disconnect $F$), and thus (!) $(E \cup F) \cap U = \emptyset$ contradicting how $U$ and $V$ were given. So $E \cup F$ is not disconnected, and hence must be connected.
