When is an inclusion map, infinitely differentiable? Let $A=\{(x,y,z)\in\mathbb{R}^3|x=2y+z\}$, which is just a line in $\mathbb{R}^3$
The question says to create an atlas on $A$ such that the inclusion map is a $C^\infty$-map.
If I understand correctly, this depends entirely on how you structure the $U_i$ in the atlas: $\{(U_i, \phi_i)\}$?
You'd want the $U_i$ to be overlapping intervals (on the line above), right?
A valid atlas would be simply $\{(A,id)\}$ with $id$ being the identity map.
Am I understanding correctly?
 A: If $X$ is a topological space, an $n$-dimensional smooth manifold structure on $X$ is a collection of open sets $U_i$ covering $X$, together with homeomorphisms $\phi_i$ of $U_i$ onto an open subset of $\mathbb{R}^n$, such that for all $i,j$ the compositions
$$\mathbb{R}^n \supseteq \phi_i(U_i \cap U_j) \rightarrow U_i \cap U_j \rightarrow \phi_j(U_i \cap U_j) \subseteq \mathbb{R}^n$$
are diffeomorphisms.  (Compatibility between charts)  The pairs $(U_i,\phi_i)$ are called charts.
Your topological space in this case is $X = \{ (x,y,z) \in \mathbb{R}^3 : x = 2y + z \}$ with the induced topology from $\mathbb{R}^3$.  You can see that $X$ is not an open subset of $\mathbb{R}^3$.  So it does not make sense to talk about the inclusion map $X \rightarrow \mathbb{R}^3$ being smooth (infinitely differentiable) without first putting a manifold structure on $X$.
Note that $X = \{ (x,y,x-2y) : x, y \in \mathbb{R} \}$.
I claim that there is a natural structure of a $2$-dimensional manifold on $X$.  Namely, take the collection of open subsets of $X$ covering $X$ to just be the open set $X$ (open in itself).    And define a map $\phi: X \rightarrow \mathbb{R}^2$ by $\phi(x,y,z) = (x,y)$.  It is easy to see that $\phi$ is a continuous bijection onto an open subset of $\mathbb{R}^2$ (namely onto the open set $\mathbb{R}^2$ itself).  Also, the inverse map $\phi^{-1}: \mathbb{R}^2 \rightarrow X$ is continuous.  So $\phi$ is a homeomorphism.
Thus we have a chart $(X,\phi)$, where $X$ is an open subset of $X$, and $\phi$ is a homeomorphism of $X$ onto $\mathbb{R}^2$.  Since we have only one chart, there is nothing to check for compatibility.  Therefore $X$, together with the single chart $(X,\phi)$, is a smooth $2$-dimensional manifold.
Let $f: X \rightarrow Y$ be a continuous function between $n$ and $m$ dimensional manifolds.  By definition, $f$ is smooth if there exist charts $(U_i,\phi_i)$ of $X$, covering $X$, and charts $(V_i,\psi_i)$ of $Y$, such that $f(U_i) \subseteq V_i$, and such that each composition 
$$\phi_i(U_i) \xrightarrow{\phi_i^{-1}} U_i \xrightarrow{f} V_i\xrightarrow{\psi_i} \psi_i(V_i)$$
is smooth as a map from an open subset of $\mathbb{R}^n$ to an open subset of $\mathbb{R}^m$.
Now $X$ and $\mathbb{R}^3$ are manifolds, so I'll leave it to you to check that the inclusion map $i: X \rightarrow \mathbb{R}^3$ is smooth.
