Show that the set $\Bbb R^2 \setminus \{-e_1, e_1\}$ is homotopy equivalent with $S(-e_1, 1) \cup S(e_1, 1)$. 
Show that the set $\Bbb R^2 \setminus \{-e_1, e_1\}$ is homotopy equivalent with $S(-e_1, 1) \cup S(e_1, 1)$.

So this is asking to show that $\Bbb R^2$ with two points missing is homotopy equivalent with the "figure eight". To show this I need to find maps $f : \Bbb R^2 \setminus \{-e_1, e_1\} \to S(-e_1, 1) \cup S(e_1, 1) $ and $g : S(-e_1, 1) \cup S(e_1, 1) \to \Bbb R^2 \setminus \{-e_1, e_1\}$ that satisfy $g \circ f \simeq id_{\Bbb R^2 \setminus \{-e_1, e_1\}}$ and $f \circ g \simeq id_{ S(-e_1, 1) \cup S(e_1, 1)}$.
I'm very confused on how to start constructing such maps. It seems as if I would need to map the whole plane to $S(-e_1, 1) \cup S(e_1, 1)$ somehow. Can I use the inclusion map here to map all points of the plane to $S(-e_1, 1) \cup S(e_1, 1)$?

You want your map g to be the inclusion map. For f you want to use the gluing lemma, to find fitting subsets of $\mathbb{R}^2$, which can be send to the figure eight (for example you can choose one of the subsets to be $B_1(-e_1)\backslash \{-e_1\}$, and find a map that gives a homotopy equivalence between $S(-e_1,1)\cong B_1(-e_1)\backslash S(-e_1,1)$
