Let $d:X\times X\rightarrow \mathbb{R}$ such that:
$d(x,y)=0 \iff x=y$
$d(y,x)\leq d(z,y)+d(z,x)$
I have started with symmetry:
Let $x=z$ so: $$d(y,x)\leq d(x,y)+d(x,x)$$
$$d(y,x)\leq d(x,y)$$
But I can find how to prove $$d(y,x)\geq d(x,y)$$ to get an equality
Inequality $(2) $ gives
$$d(y,x)\leq d(z,x)+d(z,y) $$ and $$d(x,y)\leq d(z,x)+d(z,y) $$
so with $z=x $ in the first, we get
$$d(y,x)\leq d(x,y) $$
and $z=y $ in the second gives $$d(x,y)\leq d(y,x) $$
thus $d (x,y)=d (y,x) $.
may be you will need $d (x,y)\geq 0$.
You've proved that for all $(x,y)\in X\times X$, $d(y,x)\le d(x,y)$. For any pair $(x,y)\in X\times X$, apply this inequality to the pair $(y,x)\in X\times X$. This yields the second inequality $d(x,y)\le d(y,x)$.
(2)+symmetry implies transitivity, and (1) + (2) applied to the triple $(x,x,z)$ becomes $$\forall (x,z)\in X\times X,\ 0\le 2d(z,x)$$
Which is non-negativity.
$d(y,x)\leq d(x,y)$ translated in plain English means "swapping the elements I get a distance less or equal than the previous one". If you iterate this twice you get the original distance: $$d(y,x)\leq d(x,y)\leq d(y,x)$$ which implies the equal sign.
