How to sort out different cases in a proof related to a metric space Problem:
Let $M = (X, d)$ be a metric space.
Show that 
$e(x, y) = min(1, d(x, y))$ is a metric.
To prove the triangle inequality for metric $e$,
three cases are considered.
Let $x, y, z \in X$.
(A) $d(x, y) \le 1$ and  $d(y, z) \le 1$
(B) $d(x, y) > 1$
(C) $d(y, z) > 1$
Why the above three cases are exhaustive?
Is there a systematic method to find the different cases? 

You have a metric space $X$ with a metric $d$, and you are given three points $x,y,z$. You are told to consider $d(x,y)$ and $d(y,z)$. 
Your statement simply asserts that there are only certain possibilities. There's no extra trickery here. You either have that 


*

*Both $d(x, y) \le 1$ and $d(y, z) \le 1$,

*Only one of them is less than (or equal to) $1$,

*Neither of them are less than (or equal to) $1$.

