Why is a double cone not a manifold? I'm trying to solidify my understanding of manifolds as part of a review of general relativity, and I'm having trouble understanding why a double cone isn't a manifold. I intuitively believe  most of the answers I've seen on Stack Exchange and elsewhere, which, generally speaking, point out that a neighborhood of the vertex isn't homeomorphic to any open subset of $\mathbb{R}^2$. What I'm having trouble with is that a cone seems to satisfy all the definitions I've seen for $C^\infty$ manifolds. Take this definition from Munkres' Analysis on Manifolds, for example (omitting some clarifications but otherwise copied word-for-word from pages 345-346):

"Let $M$ be a metric space. Suppose there is a collection of homeomorphisms $\alpha_i:U_i\rightarrow V_i$, where $U_i$ is open in $\mathbb{H}^k$ or $\mathbb{R}^k$, and $V_i$ is open in $M$, such that the sets $V_i$ cover $M$. Suppose that the maps $\alpha_i$ overlap with class $C^\infty$; this means that the transition function $\alpha_i^{-1} \circ \alpha_j$ is of class $C^\infty$ whenever $V_i\cap V_j$ is nonempty. The metric space $M$, together with this collection of coordinate patches on $M$, is called a differentiable k-manifold (of class $C^\infty$)."

Consider the double cone $S = \{(x,y,z)\in \mathbb{R}^3: z^2 = x^2+y^2\}$. For the sake of definiteness, let's say we make it a metric space by giving it the standard Euclidean metric inherited from $\mathbb{R}^3$. For our open sets, let $U_1 = \mathbb{R}^2$, $V_1 = \{(x,y,z)\in S: z\leq 0\}$ (the lower half of the cone plus the vertex), $U_2 = \mathbb{R}^2\backslash \{(0,0)\}$, and $V_2 = \{(x,y,z)\in S: z>0\}$ (upper half of cone). Let $\alpha_1:U_1\rightarrow V_1$ be given by $\alpha_1(x,y,z) = (x,y)$, and let $\alpha_2:U_2\rightarrow V_2$ with $\alpha_2(x,y,z) = (x,y)$. The open sets $V_1$ and $V_2$ have no overlap ($V_1\cap V_2 = \emptyset$) so the requirement that the maps $\alpha_1$ and $\alpha_2$ overlap with class $C^\infty$ is vacuously satisfied. I can't find anything in Munkres' definition that excludes situations like this. Is my misunderstanding that any open subset $V_i$ of $M$ must be homeomorphic to an open subset of $\mathbb{R}^n$, and just because we can find some open subsets that meet the criteria in the above definition does not mean the set in question is a manifold?
Would love to hear any insights or clarifications, and thanks for reading this far.
 A: The set $V_1$ you define is not an open subset of $M$; it contains the vertex but no neighborhood of the vertex. Thus, your construction does not show that $M$ is a manifold. (You are right that both $V_1$ and $V_2$ are homeomorphic to open subsets of $\mathbb R^2$ --- $\mathbb R^2$ and $\mathbb R^2\setminus 0$, respectively.)
You are correct that, to show $M$ is a manifold, it suffices to cover $M$ with some open subsets homeomorphic to open subsets of $\mathbb H^n$ or $\mathbb R^n$ (subject to agreement on overlaps). This implies that every point on $M$ has a neighborhood homeomorphic to an open subset of $\mathbb H^n$ or $\mathbb R^n$, as follows: for any point $x \in M$, $x$ must be contained in some set $V$ in the cover. Since $V$ is open, $x$ has a neighborhood $N$ contained entirely within $V$. Then the diffeomorphism from $ $ to an open subset of $\mathbb H^n$ or $\mathbb R^n$ restricts to a diffeomorphism from $N$ to an open subset of $\mathbb H^n$ or $\mathbb R^n$.
EDIT: If we are thinking about the topology on $M$ as the metric topology induced from $\mathbb R^3$, we can show $V_1$ is not open in $M$, as follows: If $V_1$ were open in $M$, there would have to be some $\epsilon > 0$ such that $V_1$ contains an $\epsilon$-ball around the vertex (i.e., all points within distance $\epsilon$ of the vertex). But no matter how small $\epsilon$ is, any $\epsilon$-ball around the origin will contain points in $V_2$, which is disjoint from $V_1$.
