Does a continuous distance satisfy the triangle inequality?

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I am not sure what tags are most appropriate here, so any help is appreciated (I found no tags for premetrics, quasimetrics, pseudometric, etc.). My level is undergrad-master.

(TL;DR: One can probably start with the questions below, and read the definitions later.)

Let a distance function on $X$ be a function $d:X\times X\to[0,\infty]$ that satisfies:

  1. reflexive: $d(x,x)=0\quad\forall x\in X$
  2. symmetric: $d(x,y)=d(y,x)\quad\forall x,y\in X$

A distance $d$ could then further satisfy the triangle inequality, i.e.

  1. $d(x,y)\leq d(x,z)+d(z,y)\quad\forall x,y,z\in X$.

We can also define convergence of sequences for a distance, saying that $\{x_n\}_n\to x$ if $\lim_{n\to\infty}d(x_n,x)=0$. And thus we could define a distance $d$ to be continuous if for any sequences $x_n,y_n$ s.t. $x_n\to x, y_n\to y$ then $d(x_n,y_n)\to d(x,y)$.

I am then interested to see what the triangle inequality (3) gives us more precisely.

We can prove that a distance $d$ satisfying (3) is continuous:

\begin{equation} |d(x_n,y_n)-d(x,y)|=|d(x_n,y_n)-d(x_n,y)+d(x_n,y)-d(x,y)|\leq \\ |d(x_n,y_n)-d(x_n,y)|+|d(x_n,y)-d(x,y)|\leq d(y_n,y)+d(x_n,x)\to 0, \text{ as }n\to\infty. \end{equation}

Question(s): However, can we prove that a continuous distance satisfies (3)? If so (not): any hints (counterexamples)? Also, could we then somewhat informally say that what (3) gives us is precisely what is needed to talk about continuity of a distance, i.e. that (3) characterizes the notion of continuity of a distance? Am I missing something obvious here or does this reasoning make sense?

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This is a concrete instance of Example $2.2$ from The Geodesic Problem in Quasimetric Spaces by Qinglan Xia.

Define $d$ on $\mathbb{R}^2$ by $$d(x,y)=|x-y|+2|x-y|^2$$

Clearly, $d$ is continuous, nonnegative, reflexive, and symmetric.

We have $$\begin{align} d(1,3)&=2+2\cdot4=10\\ d(3,6)&=3+2\cdot9=21\\ d(1,6)&=5+2\cdot25=55 \end{align}$$

and the triangle inequality doesn't hold.