Certainly not the most elementary proof, but this one feels quite satisfying conceptually: let $(X,d)$ be a metric space and contemplate a Cauchy sequence $\{x_n\}$ with a convergent subsequence, say convergent to $L \in X$. Now consider the completion $\overline{X}$ of $X$: by definition every Cauchy sequence in $\overline{X}$ converges, so our sequence $\{x_n\}$ converges in $\overline{X}$, say to $M$. But then every subsequence also converges to $M$ and thus $M = L$. It follows that the original Cauchy sequence is convergent to $L$!
Added: Really though, the direct proof is also quite simple conceptually: you want to show that the sequence converges to $L$. For any fixed $\epsilon$ we know (i) all pairs of terms with sufficiently large index are within $\frac{\epsilon}{2}$ of each other, and (ii) there is at least one term of the sequence with sufficiently large index which is within $\frac{\epsilon}{2}$ of $L$. Apply the triangle inequality!
According to your proposed definition, the sequence $1,2,3,\dots$ would be Cauchy in $\mathbb{R}$, witnessed by the sequence of open sets $U_n = (n,\infty)$.
Edit: Let me incorporate some information from the comments to make this a more complete answer.
As another example of why your definition is unsatisfactory: In $\mathbb{R}^2$, any sequence $(a_n,0)$ of points on the $x$-axis is Cauchy, witnessed by the sequence of open sets $\mathbb{R}\times (-1/n,1/n)$.
The fact that completeness and Cauchyness are not topological properties can be formalized by the fact that there are generally many different metrics compatible with a given topology, and these different metrics can induce different notions of completeness and Cauchyness. Looked at a different way, homeomorphisms preserve all topological properties (I would take this to be the definition of a topological property), but homeomorphisms between metric spaces do not necessarily preserve completeness (see the examples in btilly's and Andreas Blass's comments).
On the other hand, the notion of completeness actually lives somewhere in between the world of topological properties and metric properties, in the sense that many different metrics can agree about which sequences are Cauchy. It turns out that they agree when they induce the same uniform structure on the space. And indeed, completeness can be defined purely in terms of the induced uniform structure, so it's really a uniform property.
There is one class of spaces in which topological property and uniform properties coincide: a compact space admits a unique uniformity. So you could call completeness a topological property for compact spaces. But in a rather trivial way, since every compact uniform space is automatically complete.
It could be worthwhile to view compactness as the proper topological version of completeness, i.e. the topological property that comes the closest to agreeing with the uniform/metric property of completeness.
Best Answer
On $(0,\infty)$, consider the usual distance $d_1$ ($d_1(x,y)=|x-y|$) and the distance $d_2$ defined by $d_2(x,y)=|\log(x)-\log(y)|$. These distances are equivalent (that is, they induce the same open sets), and therefore a sequence of elements of $(0,\infty)$ converges in $\bigl((0,\infty),d_1\bigr)$ if and only if it converges in $\bigl((0,\infty),d_2\bigr)$.
However, the sequence $\left(\frac1n\right)_{n\in\Bbb N}$ is a Cauchy sequence in $\bigl((0,\infty),d_1\bigr)$, but not in $\bigl((0,\infty),d_2\bigr)$.
So, the knowledge of the open subsets of a metric space $(X,d)$ is not enough to know its Cauchy sequences.