Does a metric-like space generate a topology if open balls are defined as $B_\sigma(X,\varepsilon)=\{y\in X; |\sigma(x,y)-\sigma(x,x)|<\varepsilon\}$? The following is a quote from the paper A. Amini-Harand: Metric-like spaces, partial metric spaces and fixed points, DOI: 10.1186/1687-1812-2012-204

Definition 2.1. A mapping $\sigma\colon X\times X\to\mathbb R^+$, where $X$ is a non-empty set, is said to be metric like on $X$ if for any $x$, $y$, $z$ the following condition hold true:
  
  
*
  
*$\sigma(x,y)=0$ $\Rightarrow$ $x=y$
  
*$\sigma(x,y)=\sigma(y,x)$
  
*$\sigma(x,z)\le \sigma(x,y)+\sigma(y,z)$
  
  
  The pair $(X,\sigma)$ is then called a metric-like space. Then a metric-like on $X$ satisfies all of the conditions of a metric except that $\sigma(x,x)$ may be positive for $x\in X$. Each metric-like $\sigma$ on $X$ generates a topology $\tau_\sigma$ on $X$ whose base is the family of open $\sigma$-balls
  $$B_\sigma(x,\varepsilon)=\{y\in X; |\sigma(x,y)-\sigma(x,x)|<\varepsilon\} \qquad\text{for all $x\in X$ and $\varepsilon>0$.}$$
Then a sequence $\{x_n\}$ in a metric-like space converges to a point $x\in X$ if and only if $\lim_{n\to\infty} \sigma(x_n,x)=\sigma(x,x)$.

Question 1. Are the above claims correct? If yes how can we show that $\{B(x,\varepsilon); x\in X, \varepsilon>0\}$ is indeed a base of topology?
Side remark: I have also found the same condition under the name metametric in the Wikipedia article on metric (current revision). The reference given there is: Väisälä, Jussi (2005), "Gromov hyperbolic spaces", Expositiones
Mathematicae, 23 (3): 187–231, doi: 10.1016/j.exmath.2005.01.010. However, the topology in this paper is defined differently. (For example, a point $x$ is isolated whenever $\sigma(x,x)>0$.)
I have experimented a bit with some finite spaces. I suppose I might have missed something, but this seems to be a counterexample to the claim that this is a base.
Example. Let $X=\{a,b,c\}$ and we define $\sigma(x,y)$ as follows:
$$
\begin{array}{c|ccc}
    & a & b & c \\\hline
  a & 1 & 1 & 2 \\
  b & 1 & 2 & 2 \\
  c & 2 & 2 & 3 
\end{array}
$$
This $\sigma$ is obviously symmetric. The implication $\sigma(x,y)=0$ $\Rightarrow$ $x=y$ is vacuously true. So it remains to check triangle inequality
$$\sigma(x,z)\le \sigma(x,y)+\sigma(y,z).$$
For any choice of $x$, $y$, $z$ we have $\sigma(x,y)+\sigma(y,z)\ge1+1=2$. So the only remaining possibility to check is $x=z=c$. To see that this is true it suffices to notice that
$$\sigma(c,c)\le2\sigma(c,x)=\sigma(c,x)+\sigma(x,c)$$
for any choice of $x$. (We get either $3\le2\cdot2$ or $3\le2\cdot3$.)
As far as I can tell, the balls defined above do not give a base. We have $B_1=B(a,1/2)=\{a,b\}$ and $B_2=B(b,1/2)=\{b,c\}$. But the intersection $B_1\cap B_2=\{b\}$ does not contain any open ball $B(x,\varepsilon)$.
Question 2. Is my counterexample wrong? Where did I make a mistake?
DISCLAIMER: This is a modification of a question was originally asked on MathOverflow: Base of topology for metric-like space. From the reopen review it seems that it is unlikely to get reopened. And based on the OP comments, they seem reluctant to post it on another site.
To me the question does not seem immediately trivial. (In fact, I think that I have a counterexample, but I might have easily missed something.) And judging by the comments on MO at least one another user shown interest in it. So I decided to post it here.
EDIT: Only after posting this I have noticed these two older posts: metric-like spaces and Metric-like space. However, I see only one deleted answer there. As far as I can tell, the deleted answer uses different definition of open ball from the one given above. (It uses open ball defined in the same way as for metric spaces.)
 A: The "scaling" of the triangle inequality seems odd. Looking at the definition of a ball, I would expect that adding a constant to an existing metric does not change things much. Yet when looking at the triangle inequality, the added constant will occur twice on the right hand side and only once on the left hand side. 
As far as I see, not only does your counterexample work, but in fact you can do even worse. Let $m:X \times X \to \mathbb{R}$ be bounded, symmetric and $m(x,x) = 0$ for all $x$. (But without the need for $m$ to satisfy the triangle inequality.)
Then we can turn $m$ into something satisfying the definition by adding a constant. Let $M_0 = \sup_{x,y} |m(x,y)|$ and set
$$\sigma(x,y) = m(x,y) +3M_0.$$
Then $\sigma(x,y) > 0$ trivially, symmetry by definition and since $2M_0 \leq \sigma(x,y) \leq 4 M_0$ we have
$$\sigma(x,z) \leq 4M_0 \leq \sigma(x,y) + \sigma(y,z).$$
However as $|\sigma(x,y) - \sigma(x,x)| = |m(x,y)-m(x,x)|=|m(x,y)|$, the balls generated by $\sigma$ are the ones generated by treating $m$ as a metric, even though the triangle inequality is not satisfied by $m$ which you in general would need for those balls to generate a topology.
Interestingly the paper you cited mentions "partial metrics" directly in the introduction, with the triangle inequality being changed into
$$\sigma(x,z) \leq \sigma(x,y) + \sigma(y,z) - \sigma(y,y)$$
which seems to be the right fix, making this trick impossible.
