Proof that a discrete metric is indeed a metric space QUESTION
Let X be any set and $d : X \times X \to \mathbf{R}$ be given by $$ d(x,y) = \begin{cases}
0,  & \text{if $x = y$} \\
1, & \text{if $x \neq y$}  \\
\end{cases}$$
Show that $d$ is a metric on $X$. 
REMARKS
I'm interested in getting to understand this question. First, I assume we're discussing what is known as the Discrete Metric here. But I'm finding it a bit "too easy" to prove the axioms. And so I get the impression that I'm doing it wrong. 
For example; when looking at the axioms:


*

*$d(x,y) \ge 0$ 

*$d(x,y) = 0 ,\text{iff}: x=y$

*$d(x,y) = d(y,x)$


Both appear to be self-explanatory. The definition of the metric space does make it clear and I find myself doing nothing more than re-stating the definition of the metric to "prove" these points.
$4.$ Triangle Inequality: $d(x,y) \le d(x,z) + d(z,y)$
This I showed by considering a number of different cases (but not all of them - should I do all possible computations of the $x=/\neq y = /\neq z$ combination?) and find this to be true in each case.
I guess all I want to know - is, is the proof of this as simple as it looks?
Thanks
 A: This already has an answer, but I'd still like to share this method! Often we feel it is "necessary" to give a direct proof of the properties which define a metric, however sometimes it is far easier and less painful to simply prove by contradiction!
Assume that 
$$d(x,y) > d(x,z) + d(y,z)$$
If $x = y$ the we have an immediate contradiction. If $x \not = y$ then $d(x,y) = 1$ then we must have $d(x,z) = 0$ and $d(y,z) = 0$ but now $x = y$ so we have a contradiction.
A: Of courses some of the cases for the triangle equality can be subsumed: For example, if $x=y$, then the left side is always zero and therefore smaller than the right side. And if $x\ne y$, then all you have to show is that there is a pair of distinct values on the right side, i.e. at least one of $x\ne z,\ z\ne y$ holds.
A: We want to prove $$ d(x,y)≤d(x,z)+d(z,y) $$
Since each of the $3$ terms has two possible states ($0$ or $1$) there are $2^3=8$
  possible cases. However, we can do better using some logic. In fact we will only need to consider 2 cases: $x=y, x \neq y$
Case 1: $x=y$: By (1): the non-negativity property $$d(x,z)+d(z,y) \geq 0 $$ Therefore when $x=y$, $d(x,y)=0$ and the triangle inequality is satisfied. There's 4 cases gone. 
Case 2: $x\neq y$: In the case the only way the triangle inequality can fail is if $x=z,z=y$ giving $0\geq1$. Suppose this is possible. By the transitive property (of the equivalence relation known as equality "="):$$x=z,z=y \implies x=y$$ which is a contradiction to our assumption $x \neq y$. Therefore at least one of the following is true $x\neq z,z\neq y$. This takes care of the final 4 cases.
Therefore the inequality is satisfied.  
