How is the Knaster–Kuratowski fan connected? If the Knaster–Kuratowski fan is totally disconnected with its dispersion point removed, then how is it connected with its dispersion point in place?
How can the connectedness of the points along the lines between the Cantor set and the dispersion point rely on the dispersion point itself?
It seems to me that if the Knaster–Kuratowski fan is connected, then after the  removal of the dispersion point, you'd be left with an infinite set of lines that are themselves internally connected, one for each point in the cantor set.
 A: Well, ultimately the issue is that you aren't using the definition of connectedness.  You're using your own vague intuition about how it's supposed to behave, but that's not how connectedness is actually defined.  If you want to understand why a space is connected or not, you have to actually use the definition.
The proof that the Knaster-Kuratowski fan is connected is fairly complicated, but here's a much simpler example that shows your intuition can't be right.  Let $K$ be the Knaster-Kuratowski fan with its dispersion point removed, and consider the space $X=K\cup\{\infty\}$ with the topology that a set $U\subseteq X$ is open iff either $U=X$ or $U$ is an open subset of $K$.  Then $X$ is connected, since the only open set containing $\infty$ is all of $X$.  But if you remove $\infty$ from $X$, you're left with $K$ with its usual topology, which is totally disconnected.
As for why $K$ is totally disconnected, keep in mind that the definition of totally disconnected is not "any two points can be separated by clopen sets".  Rather, it is "no subset with more than one point is connected".  So even though you can't separate two points along the same line of $K$ by clopen subsets of $K$, they are still "disconnected" from each other in $K$ since there is no connected subset of $K$ that contains both of them.
