Why doesn't the concept of open ball imply a metric in the usual topology? In Wikipedia

One may talk about balls in any topological space X, not necessarily induced by a metric.

However, in the usual topology on R has as basis the set of all open balls, and open balls are defined as
$$B(p;r) =\{x\in X \;\vert\;d(x,p)<r\}$$
Aren't $d$ and $r$ distances, implicitly implying that a metric has been defined? In other words, how can this standard or usual topology in, for instance $\mathbb R^3,$ be properly defined as open balls without acknowledging the use of a metric?
If, on the other hand, the topology is defined as neighborhoods, the problem may not be there...
 A: 
Aren't $d$ and $r$ distances, implicitly implying that a metric has been defined?

Yes.

In other words, how can this standard or usual topology in, for instance $\mathbb R^3,$ be properly defined as open balls without acknowledging the use of a metric?

That cannot be done. You need the concept of a metric to define a ball. But that does not necessarily mean that $X$ has to be a metric space.
Actually if you read the wiki article to the end it should be clearer:

One may talk about balls in any topological space X, not necessarily induced by a metric. An (open or closed) n-dimensional topological ball of X is any subset of X which is homeomorphic to an (open or closed) Euclidean n-ball. Topological n-balls are important in combinatorial topology, as the building blocks of cell complexes.

The notion of a ball does not exist outside of the metric world. You have to touch it at some point. What wiki is saying is that there is a thing called a topological ball which is simply a subset of a topological space that is homeomorphic to an Euclidean ball. But for that you still need a notion of a (Euclidean) metric.
For example consider
$$f:[0,1]\to\mathbb{R}^2$$
$$f(t)=(\sin(t), \cos(t))$$
The image of $f$ is an arc (a piece of a circle). This function is a homeomorphism so the image $f([0,1])$ is a topological closed ball (because $[0,1]$ is a ball in $\mathbb{R}$) even though it is not a ball (because it is a subset of the standard sphere $S^1$). 
A: This is indeed a metric topology: it is induced by a metric, in the sense that it is the coarsest topology on $\mathbb R$ making the euclidian metric $d:\mathbb R \times \mathbb R \to \mathbb R_{+}$ continuous.
There are topologies not induced by metrics, and worse there are some that are not metrizable. For example, there is a topology on $\mathbb R$ that consists of itself and the empty set. This is not induced by any metric.
