How to show that two equivalence classes are either equal or have an empty intersection? For $x \in X$, let $[x]$ be the set $[x] = \{a \in X | \ x \sim a\}$.  
Show that given two elements $x,y \in X$, either
a) $[x]=[y]$ or
b) $[x] \cap [y] = \varnothing$.
How I started it is, if $[x] \cap [y]$ is not empty, then $[x]=[y]$, but then I am kind of lost. 

Once you proved that $[x] \cap [y] \neq \varnothing$ implies $[x] = [y]$ then you are done. To do so, just notice that if the intersection is not empty (say it contains $a$), then any element in $[x]$ is equivalent to $a$, and so is any element in $[y]$, so you get your result by transitivity ($[x] \subset [y]$ and $[y] \subset [x]$).
