How to prove it's not a manifold? In $R^{3}$ , let $Y_{r}$ be the set of points at distance $r>0$ from the circle $C= \{ \left(x,y,z\right) ; x^2+y^2=1,z=0 \}$ i.e. a doughnut which may be too fat.
How to prove that when $r \geq 1$, $Y_{r}$ is not a smooth manifold?
It is intuitively correct since the point above origin is not $smooth$ there.
But if I directly parameterise it and show that it can't be differentiated, it only shows that this parameterisation does not work, how to prove in general?
 A: Here's a sketch (for completeness, I added the case $r = 1$ per @CYC's comment):
Case 1 (r > 1)
At the point $p$ above the origin, you can show that there are $3$ linearly independent tangent directions, $2$ from one "sheet" of the parameterization and $1$ from the other. So, the tangent space $T_p Y_r$, and thus $Y_r$ itself, has dimension at least three. On the other hand, you can realize $Y_r$ as the level set of a function $\mathbb{R}^3 \to \mathbb{R}$ with full rank at most points on $Y_r$, so if $Y_r$ were a manifold, it would have dimension $2$, a contradiction.
Case 2 (r = 1)
In this case, the only problem point is the origin $0$. By the same argument as above, $Y_1$ must be a $2$-manifold. So, pick any local homeomorphism $\phi$ from a neighborhood $U$ of $0$ in $Y_1$ to $\mathbb{R}^2$; we may assume $U$ is connected and (by intersection) contained in a ball of radius $< 2$ centered at $0$. Then, $U - \{0\}$ is disconnected but its homeomorphic image $\phi(U - \{0\})$ is connected, a contradiction.
