Let $X\subset\mathbb{R}^3$ be the union of the coordinate axies, I want to show that $\mathbb{R}^3-X$ is homotopy equivalent to a graph, and the question asks further "which graph"
Let $X\subset\mathbb{R}^4$ be the union of the $xy$-plane and the $zw$-plane, I wish to show that $\mathbb{R}^4$ is homotopy equivalent to a surface, the question says "which surface" and I have no idea.
These are some extra questions that were left, I'll be honest I have little idea. For the first one perhaps the graph $z=\frac{1}{xy}$?
What do I know?
I am happy (confident) and have shown that $\mathbb{R}^2-\mathbb{R}$ (the x axis) is homotopy equiv to $\mathbb{S}^0$ - and retracting it to these two points.
I am answering your first question. Executive summary: there's a (particularly nice type of) homotopy equivalence between three-space with axes removed and the cube graph.
Claim 1: Let $V = \{(x,0,0)\} \cup \{(0,y,0)\} \cup \{(0,0,z)\}$ (three coordinate axes) and let $W = \{(\pm 1,0,0)\} \cup \{(0,\pm 1,0)\} \cup \{(0,0,\pm 1)\} \subset V$ (six vertices of regular octahedron). Then, $\Bbb{S}^2 \setminus W$ is a (strong) deformation retract of $\Bbb{R}^3 \setminus V$.
Define $F: (\Bbb{R}^3 \setminus V) \times [0,1] \to \Bbb{R}^3 \setminus V$ by $$ F(\mathbf{x}, t) = \biggl(\! (1-t) + \frac{t}{\|\mathbf{x}\|} \biggr) \mathbf{x}, $$ and check that for all $\mathbf{x} \in \Bbb{R}^3 \setminus V$, for all $t \in [0,1]$, and for all $\mathbf{y} \in \Bbb{S}^2 \setminus W$, $$ F(\mathbf{x}, 0) = \mathbf{x}, \quad F(\mathbf{x}, 1) \in \Bbb{S}^2 \setminus W, \quad F(\mathbf{y}, t) = \mathbf{y}. $$
Claim 2: The space $\Bbb{S}^2 \setminus W$ strongly deformation retracts onto the cube graph with eight vertices at $$ \biggl( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}} \biggr). $$
In order to wrap your head around this, I highly recommend that you draw a sketch of the coordinate axes in $3$-space and the dual cube graph that has an axis puncturing the center of each square face.