Metric Space non-negative A common proof of the non-negativity of a distance in a metric space is the following (I found one like this on Wikipedia):

$d(x,y)+d(y,x) ≥ d(x,x)$

$d(x,y)+d(x,y) ≥ 0$

$2d(x,y) ≥ 0$

$d(x,y) ≥ 0$

However, the very first step seems to assume non-negativity. If we don't assume non-negativity then it is possible that 

$d(x,y)+d(y,x) &lt; 0$

It seems wrong that we assume the conclusion in the first step of the proof...

Thanks in advance.
The first step is triangle in equality. [Put $z=x$ in $d(z,x) \leq d(z,y)+d(y,x)$ ]. If you assume that all properties of a metric except non-negativity are satisfied then  this proof does show that $d(x,y) \geq 0$ for all $x,y$. 

The second step uses the fact that $d(x,x)=0$. The third one uses symmetry: $d(x,y)=d(y,x)$.
The distance in a space $X$ is, by definition, a function $d:X \mathrm{x} X\to\mathbb{R}^+$. So the non-negativity of $d$ is given in the definition itself.