Completeness in metric spaces is somewhat a 'universal' property

If $X$ is complete, and if $Y$ is a metric space and $f : X \rightarrow Y$ is a continuous map, let $(x_n)_{n}$ be a Cauchy sequence in $X$. Then it is convergent. So the sequence $(f(x_n))_{n}$ is also convergent, and so it is Cauchy.

For the other direction, I use an idea from Mindlack's comment.

Let $X$ be a metric space. Let $(x_n)_{n}$ be a Cauchy sequence, and assume it doesn't have a limit. Consider the completion of $X$, denoted by $\overline{X}$, and the canonical isometric embedding $j : X \rightarrow \overline{X}$. Let $x_\infty \in \overline{X}$ be the limit of $(j(x_n))_n$.

Let $g : x \mapsto d_{\overline{X}}(x_\infty,x)$. It is a continuous function, and it is everywhere non-zero, by assumption. So $f := x \mapsto g(x)^{-1}$ is continuous on $X$.

Now, $(f(x_n))_n$ goes to infinity, and is Cauchy by assumption. This is a contradiction. So $(x_n)_n$ is convergent.