Proving that if a function is a metric then it is symmetric and non negative I am trying to prove that given a metric d using only the properties that it $d(a,b)=0 iff a=b$ and $d(a,c)\le d(a,b)+d(b,c)$ that  $d(a,b)=d(b,a)$ and $d(a,b) \gt 0$ I understand that it is part of the definition in most texts but it is left as an exercise in mine and I can not figure it out.
I edited my original question. I stated wrong we do not have that $d(a,a)=0$ I have that $d(a,b)=0 iff a=b$

Note that $d:\mathbb{R}\times\mathbb{R}\to \mathbb{R}$ defined as $d(a,b)=0$ is such that : 


*

*$\forall a\in\mathbb{R}, d(a,a)=0,$

*$\forall a,b,c\in\mathbb{R},d(a,c)=0\leq d(a,b)+d(a,c)=0$
but you have not that $a\neq b\implies d(a,b)>0,$ so this point should be an axiom.
For the symmetry part, consider $d':\{0,1\}\to\mathbb{R}$ defined as :
$$d'(0,0)=0,d'(1,1)=0,d'(1,0)=1 \text{ and } d'(0,1)=2.$$
Now note that :


*

*$\forall a\in\{0,1\},d'(a,a)=0,$

*$\forall a,b,c\in\{0,1\},d'(a,c)\leq d'(a,b)+d'(b,c)$ 
but $d'(0,1)=2\neq 1=d'(1,0),$ so that point should be an axiom too.
