Proving the Hausdorff property for $\kappa$-metric spaces I want to prove that the Hausdorff property holds for all $\kappa$-metric spaces.

For $\kappa \neq 1$, $(X,d)$ is a $\kappa$-metric space if $X$ is a set and $d$ is a function $X\times X \rightarrow \mathbb R$
  such that for every $x,y,z \in X$ 
$1$. $d(x,y)\ge 0$
$2$. $d(x,y)=d(y,x)$
$3$. $d(x,y)=0 \iff x=y$
$4$. $d(x,z)\le \kappa [d(x,y)+d(y,z)]$.

We put a topology on $X$ by saying $U\subseteq X$ is open iff for each $x\in U$, there exists an $\epsilon>0$ such that $B_d(x,\epsilon)\subseteq U$.
In my attempted proof, take two distinct points $x,y\in X$ such that $d(x,y)=\epsilon$.
Take the open balls $B_d\left(x,{\epsilon\over {3\kappa}}\right)$ and $B_d\left(y,{\epsilon\over {3\kappa}}\right)$. Then it can be shown that $$B_d\left(x,{\epsilon\over {3\kappa}}\right)\cap B_d\left(y,{\epsilon\over {3\kappa}}\right)=\emptyset.$$
For say there is $z\in B_d\left(x,{\epsilon\over {3\kappa}}\right)\cap B_d\left(y,{\epsilon\over {3\kappa}}\right)$. Then $$d(x,y)\le \kappa \left[{\epsilon\over {3\kappa}}+{\epsilon\over {3\kappa}}\right]\\={2\over 3}{\epsilon}\\\lt \epsilon$$ which gives a contradiction.
So, we have found two disjoint open balls in $X$ that do not intersect. 
At this  point I was thinking that Hausdorff Property has been proved for these spaces, but then I remembered that open balls in $\kappa$-metric spaces area not necessarily open sets.  And for a space to be Hausdorff , we need to find, for any two distinct points, two disjoint  open sets each containing one of them.
So the above proof of Hausdorff property in $\kappa$ metric space is wrong.
Please help me prove this.
 A: You can soup up your idea to inductively build open sets as follows.  Fix $x\neq y$; we will define two sequences of sets $U_0\subseteq U_1\subseteq U_2\subseteq\dots$ and $V_0\subseteq V_1\subseteq V_2\subseteq\dots$ by induction.  These sets will have the property that for each $n$, $d(U_n,V_n)>0$ (where $d(U_n,V_n)=\inf\{d(p,q):p\in U_n, q\in V_n\}$).
We start with $U_0=\{x\}$ and $V_0=\{y\}$.  Given $U_n$ and $V_n$, let $\epsilon=d(U_n,V_n)$, and define $$U_{n+1}=\bigcup_{p\in U_n} B_d(p,\epsilon/3\kappa^2)$$ and $$V_{n+1}=\bigcup_{q\in B_n} B_d(q,\epsilon/3\kappa^2).$$
We must show that $d(U_{n+1},V_{n+1})>0$; let $r\in U_{n+1}$ and $s\in V_{n+1}$.  Then there are $p\in U_n$ and $q\in V_n$ such that $d(p,r)<\epsilon/3\kappa^2$ and $d(s,q)<\epsilon/3\kappa^2$.  We then have $$d(p,q)\leq \kappa^2(d(p,r)+d(r,s)+d(s,q))<\frac{2\epsilon}{3}+\kappa^2d(r,s).$$
But $d(p,q)\geq d(U_n,V_n)=\epsilon$, so this gives us $d(r,s)>\epsilon/3\kappa^2$.  Thus $d(U_{n+1},V_{n+1})\geq \epsilon/3\kappa^2>0$.
Now let $U=\bigcup U_n$ and $V=\bigcup V_n$.  Then $U\cap V=\emptyset$, since $U_n\cap V_n=\emptyset$ and the sequences $(U_n)$ and $(V_n)$ are ascending.  Also, $x\in U_0\subseteq U$ and $y\in V_0\subseteq V$.  Finally, $U$ and $V$ are open, since for any $p\in U$, $p\in U_n$ for some $n$, and then $U_{n+1}$ contains a ball around $p$ (and similarly for $V$).  Thus $U$ and $V$ are disjoint open sets containing $x$ and $y$, so $X$ is Hausdorff.
More generally, this argument shows that if $A,B\subseteq X$ and $d(A,B)>0$, then $A$ and $B$ can be separated by open sets (just take $U_0=A$ and $V_0=B$).
