Criteria for metric on a set Let $X$ be a set and $d: X \times X \to X$ be a function such that $d(a,b)=0$ if and only if $a=b$. 

Suppose further that $d(a,b) ≤ d(z,a)+d(z,b)$ for all $a,b,z \in X$. 

Show that $d$ is a metric on $X$.
Let $X$ be a set and $d: X \times X \to X$ be a function such that $$d(a,b)=0\text{ if and only if}\;\; a=b,\text{ and}\tag{1}$$ $$d(a,b) ≤ d(z,a)+d(z,b)\forall a,b,z \in X.\tag{2}$$ 

There's additional criterion that needs to be met for a function $d$ to be a metric on $X$: 


You must have that $d(a, b) = d(b,a)$ for all $a, b \in X$ (symmetry).You can use the two properties you have been given to prove this.  $d(a,b)\leq d(b,a)+d(b,b)= d(b, a) + 0 = d(b,a)$ and vice versa, hence we get equality.
Having proven symmetry, you will then have that  $d(a,b) \leq d(z,a) + d(z, b) \iff d(a, b) \leq d(a, z) + d(z, b)$.
Finally, using the property immediately above, along with the $(1)$, you can establish that for all $a, b\in X$ such that $a\neq b$, we must have $d(a, b) &gt; 0$. 


Then you are done.
This must be the discrete metric.
1) The first condition follows by definition that d(a,b)=0 iff a=b;
2) Symmetry: this is trivial, because if a=b you have d(a,b)=0 and b=a gives you d(b,a)=0;
3) The triangle inequality: from the symmetry you can write d(z,a)=d(a,z).
Consider two cases:
 - if a=b, clear
 - if a is not equal to b, then either a not equal to z or z not equal to b. There you have $1\leqslant 1$ or $1\leqslant 2$.