Proving $ \mathbb{R}^3 \setminus \{(0,y,z) : y,z \in \mathbb{R}\} $ is not path connected

Question

I have my intuitions as to why this is true, but I'm having trouble formalising them, we have been given that $\mathbb{R}\setminus\{0\}$ is not path-connected, previous parts of this same question have had us prove that $\mathbb{R}^3\setminus\{(0,0,0)\}$ and $\mathbb{R}^3\setminus\{(0,0,z):z\in\mathbb{R}\}$ are path-connected.

Intuition 1: "The Lasagne Intuition". This space is, essentially, infinitely many sheets of $\mathbb{R}\setminus\{0\}$ stacked on top of each other and so it is the union of infinite not-path connected spaces, and so is not path connected.

Intuition 2: "The Partition Intuition". This space is bisected by a plane, creating two discrete infinite spaces and a path can't go from one partition to the other without "crossing" the plane.

These are the 2 reasons I came up with while speaking with peers, but I'm getting nowhere with formalising either as I don't know if either leads me to a rigorous solution. Any input will be helpful.

Answer 1

Let $Y=\Bbb R^3\setminus\{\langle 0,y,z\rangle:y,z\in\Bbb R\}$. Let $U=\{\langle x,y,z\rangle\in\Bbb R^3:x>0\}$ and $V=\{\langle x,y,z\rangle\in\Bbb R^3:x<0\}$; then $U$ and $V$ are disjoint open subsets of $Y$ whose union is $Y$, so $Y$ is not connected and therefore cannot be path connected, since every path connected space is connected.

Answer 2

Suppose that there is a path $\gamma\colon[0,1]\longrightarrow\Bbb R^3\setminus\{(0,y,z)\mid y,z\in\Bbb R\}$ such that $\gamma(0)=(-1,0,0)$ and that $\gamma(1)=(1,0,0)$. Let $\pi\colon\Bbb R^3\longrightarrow\Bbb R$ be the map defined by $\pi(x,y,z)=x$. Then $\pi$ is continuous, and therefore $\pi\circ\gamma$ is continuous too. But $(\pi\circ\gamma)(0)=-1$ and $(\pi\circ\gamma)(1)=1$; so, by the intermediate value theorem, there is a $t\in[0,1]$ such that $(\pi\circ\gamma)(t)=0$. But this means that $\gamma(t)\in\{(0,y,z)\mid y,z\in\Bbb R\}$, which is impossible.