E(f) does not have to be connected even when $\ X\ $ is.
Example: Consider $\ S^1 := \{z\in\mathbb C : |z| = 1\}\ $ -- the unit circle; and also $\ f:S^1\rightarrow S^1\ $ such that:
$$\forall_{z\in S^1}\ f(z):= z^2$$
Then $\ E(f) = \{(u\ v)\in S^1\times S^1 : u^2=v^2\}\ $ is not conected.
REMARK: If $\ f:X\rightarrow Y\ $ is such that $\ X\ $ is connected, and $\ f^{-1}(y)\ $ is connected for every $\ y\in Y\ $, then $\ E(f)\ $ is connected
Example--just a variation of the above one: E(f) is disjoint for $\ f:\mathbb R\rightarrow\mathbb R^2 $ given by: $\ \forall_{x\in\mathbb R}\ f(x):= \exp(\imath\cdot x)$.