Does there exist another way of obtaining a topological space from a metric space equally deserving of the term "canonical"?

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.

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.

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.