Suppose you have $d$ satisfying your first two axioms, and you define $d'$ to be the inf of all possible sums $d(x,a_1)+d(a_1,a_2)+\dotsb+d(a_{n-1},y)$. This will satisfy $d'(x,x)\geq 0$ and $d'(x,y)=d'(y,x)$ and $d'(x,z)\leq d'(x,y)+d'(y,z)$, so it is a pseudometric. The topology defined by this pseudometric will be the same as the one defined by your original $d$, so you have not really gained any generality. It could happen that $d'(x,y)=0$ for some $x,y$ with $x\neq y$, in which case the topology is not Hausdorff, as Ricky mentioned.

One case you could consider is $d(x,y)=(x-y)^2$ on $\mathbb{R}$. Here $d'=0$, so the topology is indiscrete.

Alternatively, you can take $X=[0,1]^2$ and $d((x,y),(u,v))=$ usual distance when $x=u$ or $y=v$, and $d((x,y),(u,v))=42$ otherwise. In this case $d'((x,y),(u,v))=|x-u|+|y-v|$ and the resulting topology is the usual one.

UPDATE: this isn't quite right. I'll leave it up for the moment and try to fix it properly later.

Suppose you have $d$ satisfying your first two axioms, and you define $d'$ to be the inf of all possible sums $d(x,a_1)+d(a_1,a_2)+\dotsb+d(a_{n-1},y)$. This will satisfy $d'(x,x)\geq 0$ and $d'(x,y)=d'(y,x)$ and $d'(x,z)\leq d'(x,y)+d'(y,z)$, so it is a pseudometric. The topology defined by this pseudometric will be the same as the one defined by your original $d$, so you have not really gained any generality. It could happen that $d'(x,y)=0$ for some $x,y$ with $x\neq y$, in which case the topology is not Hausdorff, as Ricky mentioned.

One case you could consider is $d(x,y)=(x-y)^2$ on $\mathbb{R}$. Here $d'=0$, so the topology is indiscrete.

Alternatively, you can take $X=[0,1]^2$ and $d((x,y),(u,v))=$ usual distance when $x=u$ or $y=v$, and $d((x,y),(u,v))=42$ otherwise. In this case $d'((x,y),(u,v))=|x-u|+|y-v|$ and the resulting topology is the usual one.