Let's consider the manifold $S^1$
It is well known that we need two charts to cover this manifold.
Nonetheless, we can cover the full space using a single coordinate $\theta$ which is just the angle from the center.
Now, is this a general feature? I mean, is it always possible to have in every manifold a single coordinate set that cover points that are in different charts, just as in $S^1$?
On
By definition a coordinate system is a collection of charts $\{U_i,\phi_i\}_i$ where $U_i$ are open sets in your manifold that cover your whole manifold and $\phi_i:U_i \rightarrow \mathbb{R}^n$ are homeomorphisms and the transition functions $\phi^i \circ \phi^{-j}$ are smooth where ever defined.
In this case your only chart is $(S^1,\theta)$ and but if $\theta$ was a homeomorphism this means that $\theta(S^1)$ is an open interval in the real line (since $S^1$ is connected). Therefore you can build a homeomorphism between $S^1$ and $\mathbb{R}$ which is a contradiction. You can also apply this arguement to any $S^n$ to say they won't admit single charts.
The easiest examples of manifolds admitting single charts are obviously open sets of $\mathbb{R}^n$.
On
Perhaps you're "really" asking whether or not every $n$-manifold $M$ has $\mathbf{R}^{n}$ as a covering space, i.e., whether there's a single (smooth) map $\pi:\mathbf{R}^{n} \to M$ whose restrictions to suitably small subsets define coordinate charts on $M$.
If so, the answer is "no"; even among spheres, only $S^{1}$ has this property.
No, your "single coordinate" is only a smooth (continuous) function on $S^1-(1,0)$.