Prove that we can define manifold's structure for $1.$ An open interval $A=(0,1) $ $2.B=\{(x,y)\in R^2 | ||(x,y)||<1\}$
And that we can't define manifold's structure for $3.$ An closed interval $C=[0,1]$ $4.D=\{(x,y)\in R^2 | ||(x,y)||\leq1\}$
for 1. and 2. Can I use the fact that there is a homeomorphism $id$ which map open set A onto open set A in R and open set B onto open set B in $R^2$ and so we have system of coordinate neighborhoods defined in Hausdorff space A and B ? How to prove that we can't define manifold's structure for 3. and 4.?
1 and 2 are open subsets of Hausdorff, second countable spaces, then, 1 and 2 are Hausdorff and second countable subspaces. Being open, they are already homeomorphic to open subsets of $\mathbb{R}^n$. Thus, 1 and 2 are manifolds. Any open subspace $U$ of $\mathbb{R}^n$ has the differentiable structure $F_U=\{ (U_a \cap U, \phi_a|_{U_a \cap U}); (U_a,\phi_a) \in F\}$ where $F$ is the standard differentiable structure of $\mathbb{R}^n$.
But, take in 3 any neighborhood of $1$ or $0$. They are not homeomorphic to an open set of real line (because they are, in subspace topology, of the form [0,a) or (a,1] for a<1 ). A similar argument works for 4.