I need to show that the union of xy-plane and xz-plane, i.e. the set $S:=\lbrace (x,y,z)\in\mathbb{R}^3 : z=0 \mbox{ or } y=0\rbrace$, is not a surface. Here is my claim,
$\textbf{Claim :}$ Suppose $p$ is the point $(0,0,0)$ and $U:=S\cap B(p,\epsilon)$, where $\epsilon > 0$. Then $U$ cannot be homeomorphic to any open set of $\mathbb{R}^2$.
I need help in proving the above claim. Here are my thoughts: Suppose $U$ is homeomorphic to an open set $V$ of $\mathbb{R}^2$ via homeomorphism $f$.
Now if we remove the $x$-axis from $S$ then it will have $4$ connected components, so if we can show that removal of the image of $x$-axis from $V$ can leave at most three connected components then we are through, is it the case?
Another approach : if we remove the point $p$ from $S$, then the fundamental group of the resulting space is same as the fundamental group of $X:=\lbrace (x,y,z)\in\mathbb{R}^3 : z=0 \mbox{ and } x^2+y^2=1,\mbox{ or } y=0\mbox{ and } x^2+z^2=1\rbrace$. Can the two fundamental groups $\pi_1(X)$ and $\pi_1(V\setminus \lbrace f^{-1}(p)\rbrace )$ be same? If $V$ were an open ball then I don’t think they are same (though I don’t know the proof), but $V$ is any arbitrary open set.
We need the
Proposition: Let $S\subset\mathbb{R}^3$ a regular surface. For every ${\bf{p}}\in S$, there exists a neighbourhood $V_{\bf{p}}$ of $\bf{p}$ in $S$ such that $V_{\bf{p}}$ is the graph of a differentiable function which has one of the forms: $z=f(x,y)$, $y=g(x,z)$, $x=h(y,z)$.
(Proof can be found e.g. in DoCarmo's Differential Geometry of Curves and Surfaces, p.63)
Suppose that the surface $S=\{{(x,y,z)\in\mathbb{R}^3 \;|\; z=0 \;\vee\; y=0}\}$ of two intersecting planes is regular and take the point $\bf{0}$ in $S$. For every open ball $B_{r}({\bf{0}})\subset\mathbb{R}^3$ we have that $V_r=B_{r}({\bf{0}})\cap S$ is two open intersected discs (one is lying in plane $z=0$ and the other in the plane $y=0$). But $V_r$ can not be the graph of a differentiable function of the above forms, since every one of these functions must be injective. Q.E.D.