If a 'distance function' does not possess triangle inequality property, would the limit of a converging sequence still be unique? Let $X$ be a set and $d$ be a function such that $d:X\times X\to \mathbb{R}$ such that it satisfies positivity, that is, $d(x,y)\geq 0$ and $d(x,y)=0 \iff x=y.$ Moreover suppose it satisfies symmetry property, that is, $d(x,y)=d(y,x).$ However it does not satisfy triangle inequality.
Obviously if triangle inequality was to be satisfied then this will make $(X,d)$ a metric space and subsequently every converging sequence will have a unique limit. Hence I am just curious if this property is taken away, can there still be examples such that every converging sequence has a unique limit with respect to this function $d$?
I hope I explained my question sufficiently clear, many thanks in advance!

Let $d(x,y) = (x-y)^2$ on $\Bbb R$, which satisfies the first two axioms but not the triangle inequality, because:
$$d(0,2)=4\not≤2=d(0,1)+d(1,2)$$
however limits are still unique, in fact you have the same limits as the usual metric $d(x,y)=|x-y|$, since $(x_n-y)^2\to0$ iff $|x_n-y|\to0$ (continuity of the root on positive numbers).
There exist "metrics" failing the triangle inequality without unique limits:
Let $$d(x,y)=\begin{cases}0 & x=y \\ \frac1{|xy|}
&x\neq y\end{cases}$$
on $\Bbb N$, then you have that any unbounded sequence $x_n$ converges to any integer.
