Does there exist another way of obtaining a topological space from a metric space equally deserving of the term "canonical"? Every metric space is associated with a topological space in a canonical way. According to this source, this amounts to a full functor from the category of metric spaces with continuous maps to the category of topological spaces with continuous maps.
Is it possible that there exists another way of obtaining a topological space from a metric space that is equally deserving of the label "canonical"? Perhaps something that no one has thought of yet? To say it in a different way, is there a sense in which the aforementioned functor is unique? Lets assume that the morphisms between metric spaces are precisely the continuous maps, although an answer that considers other morphisms between metric spaces is welcome.
Now obviously this is a soft question, as I have neglected to specify what it means for a map to be deserving of the term "canonical." For this reason, let me motivate the question a little.
At some point in an introductory work on analysis, the author will define the meaning of the expression "the topological space associated with (or induced by) a metric space." I'd like to know if this definition is in some sense "the unique correct definition," or whether it is only "one of many possible."

EDIT: Lets put this another way. Obviously there are many function $\mathsf{Met} \rightarrow \mathsf{Top}$, but most of them are pretty boring. So we can restrict ourselves to functors, where $\mathsf{Met}$ and $\mathsf{Top}$ are viewed as categories (technically we need to make also specify what the morphisms of $\mathsf{Met}$ should be.) Anyway, as Martin points out, we're still going to be left with lots of "boring" functors. So I guess the question is, how do we get rid of all the "boring" ones? And once we do, is the canonical functor the only one that's left? Obviously I haven't defined "boring" so this is a very soft question.
Magma suggests the following refinement of the question: does the canonical functor satisfy suitable a universal mapping property?

Here's another angle. Suppose we run into an alien species, which studies topological spaces (and calls them topotopos, and what we would call "an open set of a toplogical space" they call "an openopen of a topotopo"). They also study metric spaces (and call them metrometros.) We send that species a message asking them about the "the openopens of a metrometro." Will their notion of open set of a metric space coincide with our notion? And if so, why?
 A: Partial answer for your 'alien' question.
If we assume that aliens also think about open sets as the complements of closed sets, and if we also agree with them that being closed means being closed under limits of sequences, and that $\lim(x_n)=x$ in metric space is defined as $\lim d(x_n,x)=0$, and they also use the same real numbers, then we will also agree on what open sets are.
A: The open ball topology is unique in the following sense. Firstly, following on ideas of Flagg one can consider a generalization of metric spaces where instead of taking values in $[0,\infty ]$ the metric takes values in a value quantale. If $Met$ stands for the category of all such metric spaces, valued in all possible value quantales, and taking morphisms to be the usual $\epsilon-\delta $ notion of continuous mappings, this category is equivalent to $Top$, the usual category of topological spaces. The equivalency is given by the usual open-ball topology, correctly interpreted to use only balls of suitably positive radius, where the radius is measured in the given value quantale. So far this is just a straightforward generalization of the usual scenario, we simply do not insist on using the specific value quantale $[0,\infty ]$. 
The open ball topology functor $Met\to Top$ above is the unique concrete equivalence between the categories. In other words, if $F\colon Met\to Top$ is a an equivalence of categories and its effect on the underlying sets is trivial (so that the functor assigns to each metric a topology on the same set), then $F$ is necessarily the open ball topology construction. 
A: Let $(X,d)$ be a metric space. Then the canonical topology $\tau_{can}$ is the coarsest topology on $X$ which makes (with the product topology) $d : X \times X \to \mathbb{R}$ continuous.
Proof: It follows from the triangle inequality that $d$ is continuous in the canonical topology (in fact short when we endow $X \times X$ with the sum metric). Conversely, if $\tau$ is a topology such that $d$ is continuous with respect to $\tau$, then in particular for every $x_0 \in X$ the map $X \cong X \times \{x_0\} \to X \times X \xrightarrow{d} \mathbb{R}$ is continuous. The preimage of $(-\infty,r)$ is the open ball of radius $r$ and center $x_0$. This shows $\tau_{can} \subseteq \tau$.
It follows that the canonical forgetful functor $\mathsf{Met} \to \mathsf{Top}$ is terminal among all functors $F : \mathsf{Met} \to \mathsf{Top}$ over $\mathsf{Set}$ such that $d : F(X) \times F(X) \to \mathbb{R}$ is continuous for all $(X,d) \in \mathsf{Met}$. The initial such functor is given by the discrete topology.
So I am pretty sure that aliens will come up with the same topology.
